How can I evaluate

∑n=1∞2n3n+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?

In general, how can I evaluate

homework sequences-and-series convergence power-series

**ANSWER:-**

No need to use Taylor Series, this can be derived in a
similar way to the formula for geometric series. Lets find a general
formula for the following sum:

Sm=∑n=1mnrn.
Notice that

Sm−rSm=−mrm+1+∑n=1mrn

=−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 anyr , but in your case we have r=13 and a factor of 23 in front of the sum. That is
∑n=1∞2n3n+1=23limm→∞m(13)m+2−(m+1)(13)m+1+(13)(1−(13))2

=23(13)(23)2=12.

We can define

Skm(r)=∑n=1mnkrn.
Then the sum above considered is S1m(r) , and the geometric series is S0m(r) . We can evaluate S2m(r) by using a similar trick, and considering S2m(r)−rS2m(r) . This will then equal a combination of S1m(r) and S0m(r) which already have formulas for.
This means that k , we could work out a formula for Skm(r) , but can we find Smk(r) in general for any k ? It turns out we can, and the formula is similar to the formula for ∑mn=1nk , and involves the Bernoulli Numbers. In particular, the denominator is (1−r)k+1 .

This equality holds for any

**Added note:**We can define

*given*a
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