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What I got figured out is that I should pair up $\frac{(-1)^{n}}{3^{n}}$ giving me $\sum_{n=1}^{\infty} (\frac{1}{3})^{n}(-2n-1)$. I know that the first part converges to $\frac{3}{4}$, but the other part is divergent. How should I proceed?

Radu
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4 Answers4

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Hint: $$\sum_{k\geq1}\left(-1\right)^{k}x^{2k}=-\frac{x^{2}}{1+x^{2}}\Rightarrow\sum_{k\geq1}\left(-1\right)^{k}x^{2k+1}=-\frac{x^{3}}{1+x^{2}} $$ $$\Rightarrow\sum_{k\geq1}\left(-1\right)^{k}\left(2k+1\right)x^{2k}=\frac{d}{dx}\left(-\frac{x^{3}}{1-x^{2}}\right) $$ then take $x=1/\sqrt{3}$.

Marco Cantarini
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  • Okay, I'm on level with everything you said there. Can you explain why this is, though? $$\sum_{k\geq1}\left(-1\right)^{k}x^{2k}=-\frac{x^{2}}{1+x^{2}}$$ – Radu Jan 08 '17 at 16:58
  • @Radu It is the sum of a geometric series. Note that in your case the series starts from $1$. https://en.wikipedia.org/wiki/Geometric_series – Marco Cantarini Jan 08 '17 at 17:00
  • Okay, I suspect that the formula comes from $\sum_{k\geq0}ar^{k}=\frac{a}{1-r}$ and if I were to take your case, r would be $x^2$, and a would be $(-1)^k$. But if exchanged into the result, it doesn't add up to what it should. – Radu Jan 08 '17 at 17:26
  • @Radu In the proof $r=-x^{2}$ and $a=1$. Again note that the series in the proof starts from $1$, not from $0$. – Marco Cantarini Jan 08 '17 at 17:27
  • I understand now. I was missing the part that the sum was starting from $0$. Thank you very much. – Radu Jan 08 '17 at 17:28
  • @Radu You're welcome. – Marco Cantarini Jan 08 '17 at 17:29
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prove by induction that your finite sum $$\sum_{n=1}^k\frac{(-1)^{n+1}(2n+1)}{3^n}=3/8\, \left( -1/3 \right) ^{k+1}+3/2\, \left( -1/3 \right) ^{k+1} \left( k+1 \right) +5/8 $$ and compute then the limit for $k$ tends to $\infty$

Dr. Sonnhard Graubner
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    But how would he conjecture that specific expression? – GuPe Jan 08 '17 at 16:40
  • i have found this sum in my old book there is a collection of such sums – Dr. Sonnhard Graubner Jan 08 '17 at 16:44
  • Irrelevant. After all these comments on some similar posts of yours, you still do not get why the infinite sum is often simpler than the finite sums? To top it all, you do not even know why the identity for the finite sums hold but "have found this sum in (your) old book there is a collection of such sums"? Sorry but this is the opposite of what mathematics is about. – Did Jan 09 '17 at 11:21
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    Enlightening ^^ – tired Jan 09 '17 at 15:48
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Express $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(2n+1)}{3^{n}}=-\sum_{n=1}^{\infty} (2n+1)\left(\frac{-1}{3}\right)^n$$ and use the formula for the arrthmetico-geometric series $$\sum_{n=0}^{\infty}(an+b)r^n=\frac{b}{1-r}+\frac{ar}{(1-r)^2}\quad (|r|<1).$$

Fernando Revilla
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$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(2n+1)}{3^{n}}=\underbrace{3\sum_{n=1}^{\infty} \left(\frac{(-1)^{n+1}(n+1)}{3^{n+1}}-\frac{(-1)^{n}n}{3^{n}}\right)}_{1}+2\sum_{n=1}^{\infty} \left(-\frac{1}{3}\right)^{n} n$$

On the other hand if $|x|<1$ then $$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$ thus $$\sum_{n=1}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}$$ in othere words $$\sum_{n=1}^{\infty}nx^{n}=\frac{x}{(1-x)^2}$$ set $x=-\frac 13$, we have $$\sum_{n=1}^{\infty}n\left(-\frac 13\right)^n =-\frac{3}{16}$$

Behrouz Maleki
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