How can we find sum of above series upto infinite terms?

I don't know how to start and just need some hint.

  • $\begingroup$ put even(-) together, odd(+) together. $\endgroup$ – chenbai Jan 28 '16 at 13:20

Note that

$$\frac{1}{1 - x} = 1 + x + x^2 + x^3 + \dots$$ $$\frac{d}{dx}\left(\frac{1}{1 - x}\right) = 1 + 2x + 3x^2 + 4x^3 + \dots$$ $$x\cdot\frac{d}{dx}\left(\frac{1}{1 - x}\right) = x + 2x^2 + 3x^3 + 4x^4 + \dots$$ $$\frac{d}{dx}\left(x\cdot\frac{d}{dx}\left(\frac{1}{1 - x}\right)\right) = 1 + 2^2x + 3^2x^2 + 4^2x^3 + \dots$$

Now determine LHS and substitute in $x = -\frac{1}{5}$ (I used Wolfram):

$$-\frac{x + 1}{(x - 1)^3} = 1 + 2^2x + 3^2x^2 + 4^2x^3 + \dots$$ $$\frac{25}{54} = 1 - \frac{2^2}{5} + \frac{3^2}{5^2} - \frac{4^2}{5^3} + \dots$$

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    $\begingroup$ Great. How did you observe this? $\endgroup$ – Mathematics Jan 28 '16 at 13:26
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    $\begingroup$ @Mathematics It's a common trick to start by differentiating the series expansion of $\frac{1}{1 - x}$ whenever there are running coefficients - if you hang around this forum you'll eventually see quite a few of these neat questions :) $\endgroup$ – Yiyuan Lee Jan 28 '16 at 13:28
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    $\begingroup$ In fact, if you are OK with using Wolfram, it can solve the entire problem with no work by you. $\endgroup$ – Jeppe Stig Nielsen Jan 28 '16 at 13:37
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    $\begingroup$ @JeppeStigNielsen No, I was more interested in logic.. $\endgroup$ – Ananya Jan 28 '16 at 13:42
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    $\begingroup$ @Ananya Naturally. My comment related to the fact that the above answer apparently "cheated" in one step and asked Wolfram. Which is fine enough if you know how you could find the same answer, in principle, without Wolfram, and you only want skip the "boring" details. However, some would regard an answer with no appeal to Wolfram as better. $\endgroup$ – Jeppe Stig Nielsen Jan 28 '16 at 14:23

Say sum to $n$ terms of the series is $S_n$.

Now we have that $$S_n=1+\sum_{r=1}^n (-1)^n\cdot\frac{(n+1)^2}{5^{n}}=1+R_n$$ where $R_n=\sum_\limits{r=1}^n (-1)^n\cdot\frac{(n+1)^2}{5^{n}}=-\frac{2^2}{5}+\sum_\limits{r=1}^n (-1)^{n+1}\cdot\frac{(n+2)^2}{5^{n+1}}$

Now $$-\frac{1}{5}\cdot R_n=\sum_{r=1}^n (-1)^{n+1}\cdot\frac{(n+1)^2}{5^{n+1}}$$

Subtracting, we get $$\frac{6}{5}\cdot R_n=-\frac{2^2}{5}+\sum_{r=1}^n (-1)^{n+1}\cdot\frac{2n+3}{5^{n+1}}=-\frac{2^2}{5}+Q_n$$

Similarly $$Q_n=\frac{1}{5}+\sum_{r=1}^n (-1)^{n+2}\cdot\frac{2n+5}{5^{n+2}}$$ and $$-\frac{1}{5}\cdot Q_n=\sum_{r=1}^n (-1)^{n+2}\cdot\frac{2n+3}{5^{n+2}}$$

Again we get that $$\frac{6}{5}Q_n=\frac{1}{5}+\sum_{r=1}^n (-1)^{n+2}\cdot\frac{2}{5^{n+2}}$$

When $n\to\infty$, we have that $$Q_\infty=\frac{5}{6}(\frac{1}{5}-\frac{1}{75})=\frac{7}{45}$$

Hence we can calculate $R_\infty$ and $S_\infty$ which comes out to be $\frac{25}{54}$.


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