How can I evaluate

∞∑n=12n3n+1?

I know the answer thanks to Wolfram Alpha, but I’m more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.In general, how can I evaluate ∞∑n=0(n+1)xn?

**Answer**

No need to use Taylor series, this can be derived in a similar way to the formula for geometric series. Let’s find a general formula for the following sum: Sm=m∑n=1nrn.

Notice that

Sm−rSm=−mrm+1+m∑n=1rn=−mrm+1+r−rm+11−r=mrm+2−(m+1)rm+1+r1−r.

Hence

Sm=mrm+2−(m+1)rm+1+r(1−r)2.

This equality holds for any r, but in your case we have r=13 and a factor of 23 in front of the sum. That is

∞∑n=12n3n+1=23limm→∞m(13)m+2−(m+1)(13)m+1+(13)(1−(13))2=23(13)(23)2=12.

**Added note:**

We can define S_m^k(r) = \sum_{n=1}^m n^k r^n. Then the sum above considered is S_m^1(r), and the geometric series is S_m^0(r). We can evaluate S_m^2(r) by using a similar trick, and considering S_m^2(r) – rS_m^2(r). This will then equal a combination of S_m^1(r) and S_m^0(r) which already have formulas for.

This means that *given* a k, we could work out a formula for S_m^k(r), but can we find S_m^k(r) in general for any k? It turns out we can, and the formula is similar to the formula for \sum_{n=1}^m n^k, and involves the Bernoulli numbers. In particular, the denominator is (1-r)^{k+1}.

**Attribution***Source : Link , Question Author : backus , Answer Author : Parcly Taxel*