An easy way to solve this limit of a sum?How can I evaluate $sum_{n=0}^infty(n+1)x^n$?$ sum_{k=1}^{infty}...
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An easy way to solve this limit of a sum?
How can I evaluate $sum_{n=0}^infty(n+1)x^n$?$ sum_{k=1}^{infty} ln{left(1 + frac{1}{4 k^2}right)}$ Computing this sumalternating series test of $sum(-1)^nfrac{sqrt{n+1}-sqrt{n}}{n}$Alternative way to solve this limit?evaluate the sum of an alternating harmonic series with a fixed limitShow that a Series DivergesIs there a way to sum up the series give below??Is there an easy way to prove that this series diverges?Evaluating the floor of a tough looking summationUnderstanding summation of infinite series by defining a new functionHow to find the $limlimits_{n toinfty} frac{2^{n^2}}{(n!)^2}$
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
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$begingroup$
$$lim _{ntoinfty}sum_{k=0}^nfrac{k+1}{10^k}$$
What I've tried is to create a function out of it in order to derivate it, but I am getting nowhere. I would like to know if there is an easier way to it (I would like a high school level method for this, if there is one).
sequences-and-series
edited 9 hours ago
Parcly Taxel
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asked 9 hours ago
Jon9Jon9
364 bronze badges
New contributor
Jon9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
$$lim _{ntoinfty}sum_{k=0}^nfrac{k+1}{10^k}$$
What I've tried is to create a function out of it in order to derivate it, but I am getting nowhere. I would like to know if there is an easier way to it (I would like a high school level method for this, if there is one).
sequences-and-series
edited 9 hours ago
Parcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
asked 9 hours ago
Jon9Jon9
364 bronze badges
New contributor
Jon9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
I'd by splitting it into the sum of $k/10^k$ and $1/10^k$, where the latter is easily computable as it is a geometric series.
$endgroup$
– Viktor Glombik
9 hours ago
3
$begingroup$
Well, if you forget the "high-school level" part (or are willing to be a bit hazy on the justification of the steps) you can always consider the function $f$ defined for $xin(-1,1)$ by $f(x) = sum_{k=0}^infty x^{k+1}$. You can compute a closed-form for it and differentiate that. You can also see that $f'(x) = sum_{k=0}^infty (k+1) x^k$, so what you want if $f'(1/10)$.
$endgroup$
– Clement C.
9 hours ago
1
$begingroup$
Possible duplicate of How can I evaluate $sum_{n=0}^infty(n+1)x^n$?
$endgroup$
– Martin R
8 hours ago
add a comment |
$begingroup$
$$lim _{ntoinfty}sum_{k=0}^nfrac{k+1}{10^k}$$
What I've tried is to create a function out of it in order to derivate it, but I am getting nowhere. I would like to know if there is an easier way to it (I would like a high school level method for this, if there is one).
sequences-and-series
edited 9 hours ago
Parcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
asked 9 hours ago
Jon9Jon9
364 bronze badges
New contributor
Jon9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$$lim _{ntoinfty}sum_{k=0}^nfrac{k+1}{10^k}$$
What I've tried is to create a function out of it in order to derivate it, but I am getting nowhere. I would like to know if there is an easier way to it (I would like a high school level method for this, if there is one).
sequences-and-series
sequences-and-series
edited 9 hours ago
Parcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
asked 9 hours ago
Jon9Jon9
364 bronze badges
New contributor
Jon9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 9 hours ago
Parcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
asked 9 hours ago
Jon9Jon9
364 bronze badges
New contributor
Jon9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 9 hours ago
Parcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
edited 9 hours ago
Parcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
edited 9 hours ago
Parcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
48.2k13 gold badges77 silver badges117 bronze badges
asked 9 hours ago
Jon9Jon9
364 bronze badges
New contributor
Jon9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
Jon9Jon9
364 bronze badges
asked 9 hours ago
Jon9Jon9
364 bronze badges
364 bronze badges
New contributor
Jon9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
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1
$begingroup$
I'd by splitting it into the sum of $k/10^k$ and $1/10^k$, where the latter is easily computable as it is a geometric series.
$endgroup$
– Viktor Glombik
9 hours ago
3
$begingroup$
Well, if you forget the "high-school level" part (or are willing to be a bit hazy on the justification of the steps) you can always consider the function $f$ defined for $xin(-1,1)$ by $f(x) = sum_{k=0}^infty x^{k+1}$. You can compute a closed-form for it and differentiate that. You can also see that $f'(x) = sum_{k=0}^infty (k+1) x^k$, so what you want if $f'(1/10)$.
$endgroup$
– Clement C.
9 hours ago
1
$begingroup$
Possible duplicate of How can I evaluate $sum_{n=0}^infty(n+1)x^n$?
$endgroup$
– Martin R
8 hours ago
add a comment |
1
$begingroup$
I'd by splitting it into the sum of $k/10^k$ and $1/10^k$, where the latter is easily computable as it is a geometric series.
$endgroup$
– Viktor Glombik
9 hours ago
3
$begingroup$
Well, if you forget the "high-school level" part (or are willing to be a bit hazy on the justification of the steps) you can always consider the function $f$ defined for $xin(-1,1)$ by $f(x) = sum_{k=0}^infty x^{k+1}$. You can compute a closed-form for it and differentiate that. You can also see that $f'(x) = sum_{k=0}^infty (k+1) x^k$, so what you want if $f'(1/10)$.
$endgroup$
– Clement C.
9 hours ago
1
$begingroup$
Possible duplicate of How can I evaluate $sum_{n=0}^infty(n+1)x^n$?
$endgroup$
– Martin R
8 hours ago
1
1
$begingroup$
I'd by splitting it into the sum of $k/10^k$ and $1/10^k$, where the latter is easily computable as it is a geometric series.
$endgroup$
– Viktor Glombik
9 hours ago
$begingroup$
I'd by splitting it into the sum of $k/10^k$ and $1/10^k$, where the latter is easily computable as it is a geometric series.
$endgroup$
– Viktor Glombik
9 hours ago
3
3
$begingroup$
Well, if you forget the "high-school level" part (or are willing to be a bit hazy on the justification of the steps) you can always consider the function $f$ defined for $xin(-1,1)$ by $f(x) = sum_{k=0}^infty x^{k+1}$. You can compute a closed-form for it and differentiate that. You can also see that $f'(x) = sum_{k=0}^infty (k+1) x^k$, so what you want if $f'(1/10)$.
$endgroup$
– Clement C.
9 hours ago
$begingroup$
Well, if you forget the "high-school level" part (or are willing to be a bit hazy on the justification of the steps) you can always consider the function $f$ defined for $xin(-1,1)$ by $f(x) = sum_{k=0}^infty x^{k+1}$. You can compute a closed-form for it and differentiate that. You can also see that $f'(x) = sum_{k=0}^infty (k+1) x^k$, so what you want if $f'(1/10)$.
$endgroup$
– Clement C.
9 hours ago
1
1
$begingroup$
Possible duplicate of How can I evaluate $sum_{n=0}^infty(n+1)x^n$?
$endgroup$
– Martin R
8 hours ago
$begingroup$
Possible duplicate of How can I evaluate $sum_{n=0}^infty(n+1)x^n$?
$endgroup$
– Martin R
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Write out the infinite sum:
$$S=frac1{10^0}+frac2{10^1}+frac3{10^2}+frac4{10^3}+dots$$
Divide by ten and subtract from $S$:
$$frac1{10}S=frac1{10^1}+frac2{10^2}+frac3{10^3}+frac4{10^4}+dots$$
$$S-frac1{10}S=frac1{10^0}+frac1{10^1}+frac1{10^2}+frac1{10^3}+dots$$
This is a geometric series, whose sum can be easily calculated:
$$frac9{10}S=frac1{1-1/10}=frac{10}9$$
$$S=frac{100}{81}$$
answered 9 hours ago
Parcly TaxelParcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
$endgroup$
add a comment |
$begingroup$
When $x|<1$ $$frac{1}{1-x}=sum_{k=0}^{infty} x^k ~~~~(1).$$ Differentiate (1) w.r.t. $x$ to get $$frac{1}{(1-x)^2}= sum_{k=0}^{infty} k x^{k-1} Rightarrow frac{x.}{(1-x)^2}= sum_{k=0}^{infty} k x^{k}~~~~(2)$$ Let us put $x=1/10$ in (1) and (2) and then add them to get $$sum_{k=0}^{infty} frac{k+1}{10^k}=frac{10}{9}+frac{10}{81}=frac{100}{81}.$$
answered 8 hours ago
Dr Zafar Ahmed DScDr Zafar Ahmed DSc
2,8093 silver badges13 bronze badges
$endgroup$
add a comment |
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2 Answers
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2 Answers
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votes
$begingroup$
Write out the infinite sum:
$$S=frac1{10^0}+frac2{10^1}+frac3{10^2}+frac4{10^3}+dots$$
Divide by ten and subtract from $S$:
$$frac1{10}S=frac1{10^1}+frac2{10^2}+frac3{10^3}+frac4{10^4}+dots$$
$$S-frac1{10}S=frac1{10^0}+frac1{10^1}+frac1{10^2}+frac1{10^3}+dots$$
This is a geometric series, whose sum can be easily calculated:
$$frac9{10}S=frac1{1-1/10}=frac{10}9$$
$$S=frac{100}{81}$$
answered 9 hours ago
Parcly TaxelParcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
$endgroup$
add a comment |
$begingroup$
Write out the infinite sum:
$$S=frac1{10^0}+frac2{10^1}+frac3{10^2}+frac4{10^3}+dots$$
Divide by ten and subtract from $S$:
$$frac1{10}S=frac1{10^1}+frac2{10^2}+frac3{10^3}+frac4{10^4}+dots$$
$$S-frac1{10}S=frac1{10^0}+frac1{10^1}+frac1{10^2}+frac1{10^3}+dots$$
This is a geometric series, whose sum can be easily calculated:
$$frac9{10}S=frac1{1-1/10}=frac{10}9$$
$$S=frac{100}{81}$$
answered 9 hours ago
Parcly TaxelParcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
$endgroup$
add a comment |
$begingroup$
Write out the infinite sum:
$$S=frac1{10^0}+frac2{10^1}+frac3{10^2}+frac4{10^3}+dots$$
Divide by ten and subtract from $S$:
$$frac1{10}S=frac1{10^1}+frac2{10^2}+frac3{10^3}+frac4{10^4}+dots$$
$$S-frac1{10}S=frac1{10^0}+frac1{10^1}+frac1{10^2}+frac1{10^3}+dots$$
This is a geometric series, whose sum can be easily calculated:
$$frac9{10}S=frac1{1-1/10}=frac{10}9$$
$$S=frac{100}{81}$$
answered 9 hours ago
Parcly TaxelParcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
$endgroup$
Write out the infinite sum:
$$S=frac1{10^0}+frac2{10^1}+frac3{10^2}+frac4{10^3}+dots$$
Divide by ten and subtract from $S$:
$$frac1{10}S=frac1{10^1}+frac2{10^2}+frac3{10^3}+frac4{10^4}+dots$$
$$S-frac1{10}S=frac1{10^0}+frac1{10^1}+frac1{10^2}+frac1{10^3}+dots$$
This is a geometric series, whose sum can be easily calculated:
$$frac9{10}S=frac1{1-1/10}=frac{10}9$$
$$S=frac{100}{81}$$
answered 9 hours ago
Parcly TaxelParcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
answered 9 hours ago
Parcly TaxelParcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
answered 9 hours ago
Parcly TaxelParcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
answered 9 hours ago
Parcly TaxelParcly Taxel
48.2k13 gold badges77 silver badges117 bronze badges
48.2k13 gold badges77 silver badges117 bronze badges
add a comment |
add a comment |
$begingroup$
When $x|<1$ $$frac{1}{1-x}=sum_{k=0}^{infty} x^k ~~~~(1).$$ Differentiate (1) w.r.t. $x$ to get $$frac{1}{(1-x)^2}= sum_{k=0}^{infty} k x^{k-1} Rightarrow frac{x.}{(1-x)^2}= sum_{k=0}^{infty} k x^{k}~~~~(2)$$ Let us put $x=1/10$ in (1) and (2) and then add them to get $$sum_{k=0}^{infty} frac{k+1}{10^k}=frac{10}{9}+frac{10}{81}=frac{100}{81}.$$
answered 8 hours ago
Dr Zafar Ahmed DScDr Zafar Ahmed DSc
2,8093 silver badges13 bronze badges
$endgroup$
add a comment |
$begingroup$
When $x|<1$ $$frac{1}{1-x}=sum_{k=0}^{infty} x^k ~~~~(1).$$ Differentiate (1) w.r.t. $x$ to get $$frac{1}{(1-x)^2}= sum_{k=0}^{infty} k x^{k-1} Rightarrow frac{x.}{(1-x)^2}= sum_{k=0}^{infty} k x^{k}~~~~(2)$$ Let us put $x=1/10$ in (1) and (2) and then add them to get $$sum_{k=0}^{infty} frac{k+1}{10^k}=frac{10}{9}+frac{10}{81}=frac{100}{81}.$$
answered 8 hours ago
Dr Zafar Ahmed DScDr Zafar Ahmed DSc
2,8093 silver badges13 bronze badges
$endgroup$
add a comment |
$begingroup$
When $x|<1$ $$frac{1}{1-x}=sum_{k=0}^{infty} x^k ~~~~(1).$$ Differentiate (1) w.r.t. $x$ to get $$frac{1}{(1-x)^2}= sum_{k=0}^{infty} k x^{k-1} Rightarrow frac{x.}{(1-x)^2}= sum_{k=0}^{infty} k x^{k}~~~~(2)$$ Let us put $x=1/10$ in (1) and (2) and then add them to get $$sum_{k=0}^{infty} frac{k+1}{10^k}=frac{10}{9}+frac{10}{81}=frac{100}{81}.$$
answered 8 hours ago
Dr Zafar Ahmed DScDr Zafar Ahmed DSc
2,8093 silver badges13 bronze badges
$endgroup$
When $x|<1$ $$frac{1}{1-x}=sum_{k=0}^{infty} x^k ~~~~(1).$$ Differentiate (1) w.r.t. $x$ to get $$frac{1}{(1-x)^2}= sum_{k=0}^{infty} k x^{k-1} Rightarrow frac{x.}{(1-x)^2}= sum_{k=0}^{infty} k x^{k}~~~~(2)$$ Let us put $x=1/10$ in (1) and (2) and then add them to get $$sum_{k=0}^{infty} frac{k+1}{10^k}=frac{10}{9}+frac{10}{81}=frac{100}{81}.$$
answered 8 hours ago
Dr Zafar Ahmed DScDr Zafar Ahmed DSc
2,8093 silver badges13 bronze badges
answered 8 hours ago
Dr Zafar Ahmed DScDr Zafar Ahmed DSc
2,8093 silver badges13 bronze badges
answered 8 hours ago
Dr Zafar Ahmed DScDr Zafar Ahmed DSc
2,8093 silver badges13 bronze badges
answered 8 hours ago
Dr Zafar Ahmed DScDr Zafar Ahmed DSc
2,8093 silver badges13 bronze badges
2,8093 silver badges13 bronze badges
add a comment |
add a comment |
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1
$begingroup$
I'd by splitting it into the sum of $k/10^k$ and $1/10^k$, where the latter is easily computable as it is a geometric series.
$endgroup$
– Viktor Glombik
9 hours ago
3
$begingroup$
Well, if you forget the "high-school level" part (or are willing to be a bit hazy on the justification of the steps) you can always consider the function $f$ defined for $xin(-1,1)$ by $f(x) = sum_{k=0}^infty x^{k+1}$. You can compute a closed-form for it and differentiate that. You can also see that $f'(x) = sum_{k=0}^infty (k+1) x^k$, so what you want if $f'(1/10)$.
$endgroup$
– Clement C.
9 hours ago
1
$begingroup$
Possible duplicate of How can I evaluate $sum_{n=0}^infty(n+1)x^n$?
$endgroup$
– Martin R
8 hours ago