2

I have:

$$\sum_{n=1}^{\infty} \frac{n^2+1}{5^n}$$

I have separated the fraction in two, the second one, $\frac{1}{5^n}$ is clearly just a geometric series and simple enough. But what about the first one? Looking through my calculus book and notes I haven't found an example of this.

Any help is appreciated, especially if there's a general method to go around finding these kinds of sums. Thanks.

Blue
  • 75,673
agaminon
  • 357
  • Answer: $\frac{23}{32}$. General form: $$\sum\limits_{i=1}^\infty \frac{n^2 + 1}{a^n} = \frac{2 a^2-a+1}{(a-1)^3}$$ – David G. Stork Jan 20 '21 at 00:29
  • 2
    @DavidG.Stork Yeah, the answer itself I already got through wolfram alpha, but I'm interested in the process. – agaminon Jan 20 '21 at 00:30
  • 4
    General tool: if $f(x)=\sum_{n=0}^\infty a_n x^n$ is a power series, then $xf'(x) = \sum_{n=0}^\infty na_n x^n$. – Greg Martin Jan 20 '21 at 00:31
  • 1
    Another approach: $(1-x)^{-3} = \sum_{n=0}^\infty \binom{n+2}{2} x^n = \sum_{n=0}^\infty \frac{n^2+3n+2}{2} \cdot x^n$; $(1-x)^{-2} = \sum_{n=0}^\infty (n+1) x^n$; and $(1-x)^{-1} = \sum_{n=0}^\infty x^n$. So, try to find a combination of $(1-x)^{-3}, (1-x)^{-2}, (1-x)^{-1}$ which makes $\sum_{n=0}^\infty (n^2+1) x^n$ and then substitute $x := \frac{1}{5}$. (And subtract the $n=0$ term.) – Daniel Schepler Jan 20 '21 at 00:43

1 Answers1

2

Let $f(x) = \sum_{n=1}^{\infty}x^{-n}=\frac{1}{x-1}$. Then, we have \begin{align} f'(x)&=-\sum_{n=1}^{\infty} nx^{-n-1}= \frac{-1}{(x-1)^2}\\ f''(x) &= \sum_{n=1}^{\infty} n(n+1)x^{-n-2} = \frac{2}{(x-1)^3}. \end{align} Therefore, $$\sum_{n=1}^{\infty}\frac{n^2+1}{5^n} = \sum_{n=1}^{\infty}\frac{n(n+1)-n+1}{5^n}=25f''(5)+5f'(5)+f(5)= \frac{23}{32}.$$

Explorer
  • 3,147