text: # Thread: Evaluating this infinite sum
  
  1. ## Evaluating this infinite sum
  
  Is it possible to evaluate this infinite sum:
  
  $\displaystyle \sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)$
  
  2. Originally Posted by acevipa
  Is it possible to evaluate this infinite sum:
  
  $\displaystyle \sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)$
  This way seems messy, but it works..
  
  Skip proving convergence.... it converges.
  
  $\displaystyle \displaystyle L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^{i-1}\left(\frac{1}{3}\right)$
  
  $\displaystyle \displaystyle 3L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^{i-1}$
  
  $\displaystyle \displaystyle 2L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^i$
  
  This form is "cleaner." Write some terms
  
  $\displaystyle 2L = \dfrac{2}{3} + 2\left(\dfrac{4}{9}\right) + 3\left(\dfrac{8}{27}\right) + 4\left(\dfrac{16}{81}\right) + \dots$
  
  Consider what happens when we multiply by 3,
  
  $\displaystyle 6L = 2 + \dfrac{2\cdot4}{3} + \dfrac{3\cdot8}{9} + \dfrac{4\cdot16}{27} + \dots$
  
  Add the two together, matching up terms with common denominators.
  
  $\displaystyle 8L = 2 + \dfrac{2+2\cdot4}{3} + \dfrac{2\cdot4+3\cdot8}{9} + \dfrac{3\cdot8+4\cdot16}{27} + \dots$
  
  $\displaystyle = 2 + 5\left(\dfrac{2}{3}\right) + 8\left(\dfrac{2}{3}\right)^2 + 11\left(\dfrac{2}{3}\right)^3 + \dots$
  
  So
  
  $\displaystyle \displaystyle 8L = 2 + \sum_{i=1}^{\infty}(3i+2)\left(\frac{2}{3}\right)^ i$
  
  $\displaystyle \displaystyle = 2 + \sum_{i=1}^{\infty}3i\left(\frac{2}{3}\right)^i + 2\sum_{i=1}^{\infty}\left(\frac{2}{3}\right)^i$
  
  Simplifying,
  
  $\displaystyle 8L = 2 + 3(2L) + 4$
  
  $\displaystyle 2L = 6$
  
  $\displaystyle L = 3$
  
  3. For $\displaystyle |x|<1$ is...
  
  $\displaystyle \displaystyle f(x)= \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$ (1)
  
  ... so that is...
  
  $\displaystyle \displaystyle f^{'} (x) = \sum_{n=1}^{\infty} n\ x^{n-1} = \frac{1}{(1-x)^{2}}$ (2)
  
  Now from (2) follows...
  
  $\displaystyle \displaystyle \sum_{n=1}^{\infty} n\ (\frac{2}{3})^{n-1}\ \frac{1}{3} = \frac{1}{3}\ f^{'} (\frac{2}{3}) = 3$ (3)
  
  Kind regards
  
  $\displaystyle \chi$ $\displaystyle \sigma$
  
  4. Originally Posted by chisigma
  For $\displaystyle |x|<1$ is...
  
  $\displaystyle \displaystyle f(x)= \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$ (1)
  
  ... so that is...
  
  $\displaystyle \displaystyle f^{'} (x) = \sum_{n=1}^{\infty} n\ x^{n-1} = \frac{1}{(1-x)^{2}}$ (2)
  
  Now from (2) follows...
  
  $\displaystyle \displaystyle \sum_{n=1}^{\infty} n\ (\frac{2}{3})^{n-1}\ \frac{1}{3} = \frac{1}{3}\ f^{'} (\frac{2}{3}) = 3$ (3)
  
  Kind regards
  
  $\displaystyle \chi$ $\displaystyle \sigma$
  Great method. Thanks
  
  5. Originally Posted by acevipa
  Great method. Thanks
  If you can follow chisigma's solution you have posted this in the wrong forum and wasted the time of those helpers who have given or may have considered giving solutions that do not use calculus.
  
  CB
  
  6. Originally Posted by acevipa
  Is it possible to evaluate this infinite sum:
  
  $\displaystyle \sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)$
  $\displaystyle \displaystyle\ S_{\infty}=\frac{1}{3}+\left(\frac{2}{3}\right)\le ft(\frac{2}{3}\right)+\left(\frac{3}{3}\right)\lef t(\frac{2}{3}\right)^2+\left(\frac{4}{3}\right)\le ft(\frac{2}{3}\right)^3+.....$
  
  $\displaystyle 3\displaystyle\ S_{\infty}-1=2\left(\frac{2}{3}\right)+3\left(\frac{2}{3}\rig ht)^2+4\left(\frac{2}{3}\right)^3+....$
  
  $\displaystyle s=2\left(\frac{2}{3}\right)+3\left(\frac{2}{3}\rig ht)^2+.....$
  
  $\displaystyle \frac{2}{3}s=2\left(\frac{2}{3}\right)^2+3\left(\f rac{2}{3}\right)^3+4\left(\frac{2}{3}\right)^4....$
  
  $\displaystyle s-\frac{2}{3}s=\frac{s}{3}=2\left(\frac{2}{3}\right) +\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\righ t)^3+.....$
  
  $\displaystyle \frac{s-2}{3}=\left(\frac{2}{3}\right)+\left(\frac{2}{3}\r ight)^2+\left(\frac{2}{3}\right)^3+....$
  
  which is purely a geometric series.
  Then use the result of this to calculate the original sum.
  
  i will use
  
  $\displaystyle \Delta f(x)=f(x+1)-f(x)$ for the difference operator
  
  and $\displaystyle \sum\limits_x f(x) = g(x)$ if and only if $\displaystyle \Delta g(x) =f(x)$ (the indefinite summation)
  
  we have the summation by parts method
  
  $\displaystyle \sum\limits_{x} g(x)\Delta f(x)=f(x).g(x)-\sum\limits_{x} f(x+1).\Delta g(x)$
  
  (its similar to the integration by parts)
  
  if we apply this method "n" times we get
  [tex]\sum\limits_{x} g(x).\Delta f(x)=[/Math]
  
  $\displaystyle \sum\limits^{n}_{k=0}\Delta^{k}g(x).(-1)^{k}\Delta^{-k}E^{k}f(x)+(-1)^{n+1}\sum\limits_{x}\Delta^{n+1}g(x)\Delta^{-n}E^{n+1}f(x)$
  
  ( where $\displaystyle E^k f(x)=f(x+k)$)
  
  if, g(x) has degree n , (so $\displaystyle \Delta ^{n+1} g(x)=0$ )and $\displaystyle |a|<1$ we can use it on the summation by parts to find
  
  $\displaystyle \sum\limits^{\infty}_{x=c} g(x)a^{x}=a^{c-1}\sum\limits^{n}_{k=0}\frac{\Delta^{k}g(x)|_{x=c} }{(b-1)^{k+1}}$
  
  where $\displaystyle b= \frac{1}{a}$
  
  if $\displaystyle g(x)=x^{n}$ and e c=0
  $\displaystyle \sum\limits^{\infty}_{x=0} x^{n}a^{x}=a^{-1}\sum\limits^{n}_{k=0}\frac{\Delta^{k}x^{n}|_{x=0 }}{(b-1)^{k+1}}$
  
  but $\displaystyle \Delta^{k}x^{n}|_{x=0}=k!\bigg\{^{n}_{k}\bigg\}$ (here $\displaystyle \bigg\{^{n}_{k}\bigg\}$ it's the stirling number of the second kind, knuth's notation )
  so
  $\displaystyle \sum\limits^{\infty}_{x=0} x^{n}a^{x}= a^{-1}\sum\limits^{n}_{k=0}\frac{k! \bigg\{^{n}_{k}\bigg\}}{(b-1)^{k+1}}.$
  
  8. (other way , the general case)
  
  we can define the operator xD, (xD) f(x) =x [D f(x)] ( where "D" is the derivative )
  then we can show
  
  $\displaystyle (xD)^{n}g(x)=\sum\limits^{n}_{k=0}\bigg\{^{n}_{k}\ bigg\}x^{k}D^{k}g(x)$
  
  (again the stirling numbers of the second kind)
  take the series
  $\displaystyle \sum\limits^{\infty}_{n=0}x^{n}=\frac{1}{1-x}$
  
  apply $\displaystyle (xD)^p$ in the both sides
  
  $\displaystyle \sum\limits^{\infty}_{n=0}n^{p}x^{n}=(xD)^p\frac{1 }{1-x}$
  
  using the identity
  $\displaystyle (xD)^{p}g(x)=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\ bigg\}x^{k}D^{k}g(x)$
  
  we have
  
  $\displaystyle (xD)^{p}\frac{1}{1-x}=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\bigg\}x^{k }D^{k}\frac{1}{1-x}=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\bigg\}\fra c{k!x^{k}}{(1-x)^{k+1}}$
  
  so
  $\displaystyle \sum\limits^{\infty}_{n=0}n^{p}x^{n}=\sum\limits^{ p}_{k=0}\bigg\{^{p}_{k}\bigg\}\frac{k!x^{k}}{(1-x)^{k+1}}.$
explanation: The document is a discussion thread on a mathematics forum, where the initial post poses a question about evaluating an infinite sum. The thread contains multiple responses providing different methods to solve the problem, including detailed step-by-step solutions. The initial question is answered multiple times by different users, and the final answer is confirmed to be 3.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Thread: Evaluating this infinite sum
  
  1. ## Evaluating this infinite sum
  
  Is it possible to evaluate this infinite sum:
  
  $\displaystyle \sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)$
  
  2. Originally Posted by acevipa
  Is it possible to evaluate this infinite sum:
  
  $\displaystyle \sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)$
  This way seems messy, but it works..
  
  Skip proving convergence.... it converges.
  
  $\displaystyle \displaystyle L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^{i-1}\left(\frac{1}{3}\right)$
  
  $\displaystyle \displaystyle 3L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^{i-1}$
  
  $\displaystyle \displaystyle 2L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^i$
  
  This form is "cleaner." Write some terms
  
  $\displaystyle 2L = \dfrac{2}{3} + 2\left(\dfrac{4}{9}\right) + 3\left(\dfrac{8}{27}\right) + 4\left(\dfrac{16}{81}\right) + \dots$
  
  Consider what happens when we multiply by 3,
  
  $\displaystyle 6L = 2 + \dfrac{2\cdot4}{3} + \dfrac{3\cdot8}{9} + \dfrac{4\cdot16}{27} + \dots$
  
  Add the two together, matching up terms with common denominators.
  
  $\displaystyle 8L = 2 + \dfrac{2+2\cdot4}{3} + \dfrac{2\cdot4+3\cdot8}{9} + \dfrac{3\cdot8+4\cdot16}{27} + \dots$
  
  $\displaystyle = 2 + 5\left(\dfrac{2}{3}\right) + 8\left(\dfrac{2}{3}\right)^2 + 11\left(\dfrac{2}{3}\right)^3 + \dots$
  
  So
  
  $\displaystyle \displaystyle 8L = 2 + \sum_{i=1}^{\infty}(3i+2)\left(\frac{2}{3}\right)^ i$
  
  $\displaystyle \displaystyle = 2 + \sum_{i=1}^{\infty}3i\left(\frac{2}{3}\right)^i + 2\sum_{i=1}^{\infty}\left(\frac{2}{3}\right)^i$
  
  Simplifying,
  
  $\displaystyle 8L = 2 + 3(2L) + 4$
  
  $\displaystyle 2L = 6$
  
  $\displaystyle L = 3$
  
  3. For $\displaystyle |x|<1$ is...
  
  $\displaystyle \displaystyle f(x)= \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$ (1)
  
  ... so that is...
  
  $\displaystyle \displaystyle f^{'} (x) = \sum_{n=1}^{\infty} n\ x^{n-1} = \frac{1}{(1-x)^{2}}$ (2)
  
  Now from (2) follows...
  
  $\displaystyle \displaystyle \sum_{n=1}^{\infty} n\ (\frac{2}{3})^{n-1}\ \frac{1}{3} = \frac{1}{3}\ f^{'} (\frac{2}{3}) = 3$ (3)
  
  Kind regards
  
  $\displaystyle \chi$ $\displaystyle \sigma$
  
  4. Originally Posted by chisigma
  For $\displaystyle |x|<1$ is...
  
  $\displaystyle \displaystyle f(x)= \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$ (1)
  
  ... so that is...
  
  $\displaystyle \displaystyle f^{'} (x) = \sum_{n=1}^{\infty} n\ x^{n-1} = \frac{1}{(1-x)^{2}}$ (2)
  
  Now from (2) follows...
  
  $\displaystyle \displaystyle \sum_{n=1}^{\infty} n\ (\frac{2}{3})^{n-1}\ \frac{1}{3} = \frac{1}{3}\ f^{'} (\frac{2}{3}) = 3$ (3)
  
  Kind regards
  
  $\displaystyle \chi$ $\displaystyle \sigma$
  Great method. Thanks
  
  5. Originally Posted by acevipa
  Great method. Thanks
  If you can follow chisigma's solution you have posted this in the wrong forum and wasted the time of those helpers who have given or may have considered giving solutions that do not use calculus.
  
  CB
  
  6. Originally Posted by acevipa
  Is it possible to evaluate this infinite sum:
  
  $\displaystyle \sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)$
  $\displaystyle \displaystyle\ S_{\infty}=\frac{1}{3}+\left(\frac{2}{3}\right)\le ft(\frac{2}{3}\right)+\left(\frac{3}{3}\right)\lef t(\frac{2}{3}\right)^2+\left(\frac{4}{3}\right)\le ft(\frac{2}{3}\right)^3+.....$
  
  $\displaystyle 3\displaystyle\ S_{\infty}-1=2\left(\frac{2}{3}\right)+3\left(\frac{2}{3}\rig ht)^2+4\left(\frac{2}{3}\right)^3+....$
  
  $\displaystyle s=2\left(\frac{2}{3}\right)+3\left(\frac{2}{3}\rig ht)^2+.....$
  
  $\displaystyle \frac{2}{3}s=2\left(\frac{2}{3}\right)^2+3\left(\f rac{2}{3}\right)^3+4\left(\frac{2}{3}\right)^4....$
  
  $\displaystyle s-\frac{2}{3}s=\frac{s}{3}=2\left(\frac{2}{3}\right) +\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\righ t)^3+.....$
  
  $\displaystyle \frac{s-2}{3}=\left(\frac{2}{3}\right)+\left(\frac{2}{3}\r ight)^2+\left(\frac{2}{3}\right)^3+....$
  
  which is purely a geometric series.
  Then use the result of this to calculate the original sum.
  
  i will use
  
  $\displaystyle \Delta f(x)=f(x+1)-f(x)$ for the difference operator
  
  and $\displaystyle \sum\limits_x f(x) = g(x)$ if and only if $\displaystyle \Delta g(x) =f(x)$ (the indefinite summation)
  
  we have the summation by parts method
  
  $\displaystyle \sum\limits_{x} g(x)\Delta f(x)=f(x).g(x)-\sum\limits_{x} f(x+1).\Delta g(x)$
  
  (its similar to the integration by parts)
  
  if we apply this method "n" times we get
  [tex]\sum\limits_{x} g(x).\Delta f(x)=[/Math]
  
  $\displaystyle \sum\limits^{n}_{k=0}\Delta^{k}g(x).(-1)^{k}\Delta^{-k}E^{k}f(x)+(-1)^{n+1}\sum\limits_{x}\Delta^{n+1}g(x)\Delta^{-n}E^{n+1}f(x)$
  
  ( where $\displaystyle E^k f(x)=f(x+k)$)
  
  if, g(x) has degree n , (so $\displaystyle \Delta ^{n+1} g(x)=0$ )and $\displaystyle |a|<1$ we can use it on the summation by parts to find
  
  $\displaystyle \sum\limits^{\infty}_{x=c} g(x)a^{x}=a^{c-1}\sum\limits^{n}_{k=0}\frac{\Delta^{k}g(x)|_{x=c} }{(b-1)^{k+1}}$
  
  where $\displaystyle b= \frac{1}{a}$
  
  if $\displaystyle g(x)=x^{n}$ and e c=0
  $\displaystyle \sum\limits^{\infty}_{x=0} x^{n}a^{x}=a^{-1}\sum\limits^{n}_{k=0}\frac{\Delta^{k}x^{n}|_{x=0 }}{(b-1)^{k+1}}$
  
  but $\displaystyle \Delta^{k}x^{n}|_{x=0}=k!\bigg\{^{n}_{k}\bigg\}$ (here $\displaystyle \bigg\{^{n}_{k}\bigg\}$ it's the stirling number of the second kind, knuth's notation )
  so
  $\displaystyle \sum\limits^{\infty}_{x=0} x^{n}a^{x}= a^{-1}\sum\limits^{n}_{k=0}\frac{k! \bigg\{^{n}_{k}\bigg\}}{(b-1)^{k+1}}.$
  
  8. (other way , the general case)
  
  we can define the operator xD, (xD) f(x) =x [D f(x)] ( where "D" is the derivative )
  then we can show
  
  $\displaystyle (xD)^{n}g(x)=\sum\limits^{n}_{k=0}\bigg\{^{n}_{k}\ bigg\}x^{k}D^{k}g(x)$
  
  (again the stirling numbers of the second kind)
  take the series
  $\displaystyle \sum\limits^{\infty}_{n=0}x^{n}=\frac{1}{1-x}$
  
  apply $\displaystyle (xD)^p$ in the both sides
  
  $\displaystyle \sum\limits^{\infty}_{n=0}n^{p}x^{n}=(xD)^p\frac{1 }{1-x}$
  
  using the identity
  $\displaystyle (xD)^{p}g(x)=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\ bigg\}x^{k}D^{k}g(x)$
  
  we have
  
  $\displaystyle (xD)^{p}\frac{1}{1-x}=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\bigg\}x^{k }D^{k}\frac{1}{1-x}=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\bigg\}\fra c{k!x^{k}}{(1-x)^{k+1}}$
  
  so
  $\displaystyle \sum\limits^{\infty}_{n=0}n^{p}x^{n}=\sum\limits^{ p}_{k=0}\bigg\{^{p}_{k}\bigg\}\frac{k!x^{k}}{(1-x)^{k+1}}.$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion thread on a mathematics forum, where the initial post poses a question about evaluating an infinite sum. The thread contains multiple responses providing different methods to solve the problem, including detailed step-by-step solutions. The initial question is answered multiple times by different users, and the final answer is confirmed to be 3.
  
  Label: question-and-answer