Find the approximate sum of the series $\sum_{n=0}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}$ Prove that the series converges and find the approximate sum. $$\sum_{n=0}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}$$
To prove that converges what I did is to use the Cauchy's convergence test.
$$\lim_{x\to \infty} \sqrt[n]{a_n} =\lim_{x\to \infty} \frac{1}{\sqrt[n]{3^{n}\sqrt{n+1}}} =\lim_{x\to \infty} \frac{1}{3\sqrt[2n]{n+1}} = \frac{1}{3\sqrt{\lim_{x\to \infty}\sqrt[n]{n+1}}} = \frac{1}{3\sqrt{\lim_{x\to \infty}\sqrt[n]{n(1+\frac{1}{n})}}} = \frac{1}{3\sqrt{\lim_{x\to \infty}\sqrt[n]{n}\sqrt[n]{(1+\frac{1}{n})}}} = \frac{1}{3} \lt 1$$
Since $\lim_{x\to \infty} \sqrt[n]{n} = 1$ and $\lim_{x\to \infty} \sqrt[n]{(1+\frac{1}{n})} = 1$
Then, the series converges. Now, I'm not sure how to find the approximate sum of the series since it's not an alternating series. Is there a unique method?
Thanks in advance.
 A: Another way to check convergence is the comparison test
$$\sum_{n=0}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}<\sum_{n=0}^{\infty} \frac{1}{3^n}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}$$
This also provides a way to approximate the sum to as many places as required. Lets say you want to approximate the sum to $\epsilon$ accuracy. Then it is sufficient to find $N$ such that
$$ \sum_{n=N}^{\infty} \frac{1}{3^n}\leq \epsilon$$
as
$$\sum_{n=N}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}<\sum_{n=N}^{\infty} \frac{1}{3^n}\leq \epsilon$$
Of course, this geometric series is easily found to be
$$\sum_{n=N}^{\infty} \frac{1}{3^n}=\frac{\frac{1}{3^N}}{1-\frac{1}{3}}=\frac{2}{3^{N-1}}$$
Solving, we get
$$N\geq \left\lceil\frac{\ln(6/\epsilon)}{\ln(3)}\right\rceil$$
(as $N$ is an integer). We conclude that if $N$ is given by the equation above, then
$$\left|\sum_{n=0}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}-\sum_{n=0}^{N-1} \frac{1}{(3^n)\sqrt{n+1}}\right|=\sum_{n=N}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}<\epsilon$$
To find your approximation, simply calculate
$$\sum_{n=0}^{N-1} \frac{1}{(3^n)\sqrt{n+1}}$$
manually.
