Tricky limitting sum

Question

Find the sum of the following infinite series, given $|x|<1$

$$2+4x+\frac{9}{2}x^2+\frac{16}{3}x^3+\frac{25}{4}x^4+\frac{36}{5}x^5+\frac{49}{6}x^6+\frac{64}{7}x^7+\frac{81}{8}x^8+ \ldots $$

I have tried turning this into a geometric series, but I just didn't even know where to begin. I also tried relating this to well-known sums but that didn't work out either.

I would love some help with this.

Answer 1

The series appears to take the form $$ f(x) = 2 + \sum_{n=1}^{\infty} \frac{(n+1)^2}{n} \, x^n $$ which can be seen in the form: \begin{align} f(x) &= 2 + \sum_{n=1}^{\infty} \frac{(n+1)(n+2)}{n} \, x^n - \sum_{n=1}^{\infty} \frac{(n+1) \, x^n}{n} \\ &= 2 + \frac{d^2}{d x^2} \, \sum_{n=1}^{\infty} \frac{x^{n+2}}{n} - \frac{d}{dx} \, \sum_{n=1}^{\infty} \frac{x^{n+1}}{n} \\ &= 2 + \frac{d^2}{d x^2} \, (- x^2 \, \ln(1-x) ) - \frac{d}{d x} \, (- x \, \ln(1-x) ) \\ &= \cdots \\ &= 2 + \frac{3 x - 2 x^2}{(1-x)^2} - \ln(1-x) \\ &= \frac{2 - x}{(1-x)^2} - \ln(1-x). \end{align}

A faster way is: \begin{align} f(x) &= 2 + \sum_{n=1}^{\infty} \frac{(n+1)^2}{n} \, x^n \\ &= 2 + \sum_{n=1}^{\infty} \frac{n (n+2) + 1}{n} \, x^n \\ &= 2 + \sum_{n=1}^{\infty} \left( (n+2) + \frac{1}{n} \right) \, x^n \\ &= \sum_{n=0}^{\infty} (n+2) \, x^n + \sum_{n=1}^{\infty} \frac{x^n}{n} \\ &= \frac{2-x}{(1-x)^2} - \ln(1-x). \end{align}

Answer 2

Notice that $\frac{(n+1)^2}{n}=n+2+\frac{1}{n}$ so the sum changes to sum of simple series.

Answer 3

So your sum looks like $$ s(x) = 2 + \sum_{k=1}^\infty \frac{(k+1)^2}{k} x^k. $$ One approach is to expand the sum to $$ \begin{split} \sum_{k=1}^\infty \frac{(k+1)^2}{k} x^k &= \sum_{k=1}^\infty \frac{k^2+2k+1}{k} x^k \\ &= \sum_{k=1}^\infty kx^k + 2 \sum_{k=1}^\infty x^k + \sum_{k=1}^\infty \frac{x^k}{k}. \end{split} $$ The first sum you can get by differentiating a geometric series: $$ \begin{split} g(x) &= \sum_{k=0}^\infty x^k\\ g'(x) &= \sum_{k=1}^\infty k x^{k-1}\\ x g'(x) &= \sum_{k=1}^\infty k x^k \end{split} $$ The second sum is a geometric series. Differentiate the third sum to get a geometric series: $$ \begin{split} f(x) &= \sum_{k=1}^\infty \frac{x^k}{k}\\ f'(x) &= \sum_{k=1}^\infty x^{k-1} \end{split} $$ which you can now find...

Just make sure to be careful about the term where $k=0$ everywhere...

Answer 4

Note that the series can be written as, $$2+\sum_{k\geq 1}\frac{(k+1)^{2}}{k}x^{k}$$ $$2+\sum_{k\geq 1}kx^{k}+\frac{x^{k}}{k}+2x^{k}$$ Provided that $|x|<1$ we have that, $$\sum_{k\geq 1}\frac{x^{k}}{k}=-\ln(1-x)$$ $$\sum_{k\geq 1}kx^{k}=\frac{x}{(1-x)^{2}}$$ And geometric series is already known.