Sum $\frac{1}{2} + 1 + \frac{9}{8}+1+\frac{25}{32}+\frac{36}{64}+\cdots$

Question

Let $\frac{1}{2} + 1 + \frac{9}{8}+1+\frac{25}{32}+\frac{36}{64}+\cdots$ be a series.
I need to calculate its sum using $$\sum_{n=1}^{\infty}n^2x^n.$$

I was trying to set $x=\frac{1}{2}$, so: $$\sum_{n=1}^\infty n^2x^n=x\sum_{n=1}^\infty n^2x^{n-1}=x\sum_{n=1}^\infty (nx^n)'=x^2\sum_{n=1}^\infty (x^n)''.$$

Then I was differentiating separately:
$$x^2\sum_{n=1}^\infty (x^n)'' = x^2\left(\frac{1}{1-x}\right)''=x^2\left(\frac{1}{(1-x)^2}\right)'=x^2\frac{2(1-x)}{(1-x)^4}=\frac{2x^2}{(1-x)^3}.$$

And finally calculate the sum:

$$\frac{2 \cdot \frac{1}{4}}{\frac{1}{2}^3} = 4.$$

But I got the wrong answer.

What did I miss? How do I solve it correctly?

You need to include $n=0$ to summation of the derivatives. It should no affect the result. — herb steinberg, Jun 12 '21 at 20:58
Going from first derivative to second is incorrect. You can't simply pull $x$ out. — herb steinberg, Jun 12 '21 at 21:14
I always find it easier to do these staring with $\sum_{n=1}^\infty = \frac x{1-x}$, differentiate both sides, multiply through by $x$, differentiate again, multiply by $x$ again. — saulspatz, Jun 12 '21 at 23:17
Does this answer your question? How to obtain the sum of the following series? $\sum_{n=1}^\infty{\frac{n^2}{2^n}}$ Found using Approach0. — Toby Mak, Jun 13 '21 at 05:15

Mathew Mahindaratne · Accepted Answer · 2021-06-13T04:10:41.567

Let $\frac{1}{2} + 1 + \frac{9}{8}+1+\frac{25}{32}+\frac{36}{64}+\cdots$ be a series.
Calculate its sum using following relation. $$\sum_{n=1}^{\infty}n^2x^n$$

Consider following sum: $$\sum_{n=1}^{\infty}x^n = \frac{x}{1-x} \tag1$$ where $x \lt 1$ and $x \ne 0$. Differentiate both sides of the equation $(1)$: $$\sum_{n=1}^{\infty} nx^{n-1} = \frac{(1-x)-x(-1)}{(1-x)^2}= \frac{1}{(1-x)^2} \ \Rightarrow \ \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2} \tag2$$ Again, differentiate both sides of the equation $(2)$: $$\sum_{n=1}^{\infty} n^2x^{n-1} = \frac{(1-x)^2 - 2x(1-x)(-1)}{(1-x)^4} = \frac{(1-x)(1+ x)}{(1-x)^4} = \frac{(1+ x)}{(1-x)^3} $$

$$\therefore \ \sum_{n=1}^{\infty} n^2x^n = \frac{x(1+ x)}{(1-x)^3} $$

In your case, apply $x = \frac12$ and you get the solution as $6$:

$$\sum_{n=1}^{\infty} n^2\left(\frac12\right)^n = \frac{\frac12(1+ \frac12)}{(1-\frac12)^3} = 6$$

@ Lalit Tolani: I corrected it. Thought $n$ is a constant for some reason! :-) — Mathew Mahindaratne, Jun 13 '21 at 04:12

score 0 · Answer 2 · answered Jun 13 '21 at 07:51

Hint

Use the samml trick

$$\sum_{n=1}^{\infty}n^2x^n=\sum_{n=1}^{\infty}[n(n-1)+n]x^n=\sum_{n=1}^{\infty}n(n-1)x^n+\sum_{n=1}^{\infty}nx^n$$ $$\sum_{n=1}^{\infty}n^2x^n=x^2\sum_{n=1}^{\infty}n(n-1)x^{n-2}+x\sum_{n=1}^{\infty}nx^{n-1}$$ $$\sum_{n=1}^{\infty}n^2x^n=x^2\left(\sum_{n=1}^{\infty}x^n \right)''+x\left(\sum_{n=1}^{\infty}x^n \right)'$$

score 0 · Answer 3 · answered Jun 13 '21 at 08:09

0

$$ \sum_{k=1}^{\infty}k^2x^k = x\left(\sum_{k=1}^{\infty}kx^k\right)'=x\left(x\left(\sum_{k=1}^{\infty}x^k\right)'\right)' = x\left(x\left(\frac{1}{1-x}\right)'\right)'=\frac{x (x+1)}{(1-x)^3} $$

answered Jun 13 '21 at 08:09

Cesareo

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Sum $\frac{1}{2} + 1 + \frac{9}{8}+1+\frac{25}{32}+\frac{36}{64}+\cdots$

3 Answers3