Evaluate the infinite sum of $\sum_{n=0}^{\infty}\frac{1}{3^n(n+1)}$

Question

Would someone be able to show which well-known power series works with this? And help evaluate it? The sum is $\sum_{n=0}^{\infty}\frac{1}{3^n(n+1)}$.

Answer 1

Use the geometric series:

$$ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$

Now, to get $ \frac{1}{i+1}$ in denominator sum, we can get that factor by integrating both side w.r.t.x

$$ -\ln(1-x)= \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1 } +C$$

Now, we need put $x=\text{something}$ to make this sum match with the one in the question, I'll let you finish the problem.

Note: I've used the property that you can interchange integrals and summations in the domain of summation in the integration step.

Answer 2

Consider the classic series $$ \sum_{n=0}^{\infty} t^{n} = \frac{1}{1-t} $$ and integrate both sides. This leads to \begin{align} \int_{0}^{t} \sum_{n=0}^{\infty} u^n \, du &= \int_{0}^{t} \frac{du}{1-u} \\ \sum_{n=0}^{\infty} \left[ \frac{u^{n+1}}{n+1} \right]_{0}^{t} &= \left[ - \ln(1-u) \right]_{0}^{t} \\ \sum_{n=0}^{\infty} \frac{t^{n+1}}{n+1} &= - \ln(1-t) \end{align} and finally $$ \sum_{n=0}^{\infty} \frac{t^n}{n+1} = - \frac{\ln(1-t)}{t}. $$

Letting $t$ be a value within the range of convergence gives the desired series.