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I'm a little confused about how to find sums. I know that to find the sum, we need to take $\lim_{n\ \to \infty} s_n$, where $s_n$ is the partial sum. This is confusing for me b/c I have trouble telling the difference between $s_n$ and $a_n$, where $a_n$ is the series formula in $\sum a_n$.

So anyway, this is my approach:

$$s_n = \sum_{i=1}^n \frac{3^{i+1}}{6^i+1} = \frac{9}{7} + \frac{27}{37} + \frac{81}{217} + ...+ \frac{3^{n+1}}{6^n+1}$$

See, I know ultimately that I have to take the limit of $s_n$, however any convergent series (I know this converges) will have a limit = 0. So i'm assuming that in order to find the sum I need to manipulate the sum formula. Is this the right approach?

Thanks!

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    $\begingroup$ Are you sure there's a closed-form solution? Or were you just asked to show that the series converges? $\endgroup$ – Robert Israel Dec 11 '16 at 19:21
  • $\begingroup$ @RobertIsrael I was asked to show that it converges. And did so by comparison with $(\frac{1}{2})^n$ $\endgroup$ – Romaion Dec 11 '16 at 19:25
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    $\begingroup$ Then why are you trying to find the sum? Perhaps you're under the false impression that if it's easy to show that a series converges, it must also be easy to find its value $\endgroup$ – Omnomnomnom Dec 11 '16 at 19:26
  • $\begingroup$ Then you answered the question. I don't think there's any more to do with this one. $\endgroup$ – Robert Israel Dec 11 '16 at 19:26
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    $\begingroup$ The sum exists. It's approximately $2.76355818968079$. But AFAIK there isn't a closed-form formula for it. $\endgroup$ – Robert Israel Dec 11 '16 at 19:29
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$$ \frac{3^n}{6^n+1}=\frac{3^n}{6^n(1+6^{-n})}=\frac{1}{2^n(1+6^{-n})}=\frac{A}{2^n}-\frac{B}{1+6^{-n}}=\frac{1}{2^n}-\frac{12^{-n}}{1+6^{-n}}= \\[6mm] \frac{1}{2^n}-\frac{1}{6^n(6^n+1)}=\frac{1}{2^n}-\frac{1}{6^n}+\frac{1}{6^n+1}\quad\Rightarrow\quad\sum_{n=1}^{\infty}\frac{3^{n+1}}{6^n+1}=\\[6mm] 3\sum_{n=1}^{\infty}\left(\frac{1}{2^n}-\frac{1}{6^n}+\frac{1}{6^n+1}\right)=\color{red}{3\left(1-\frac{1}{5}+\frac{1}{\log6}\left[\varPsi_{1/6}\left(1-\frac{i\,\pi}{\log6}\right)-\log\frac65\right]\right)} \\[8mm] \small\qquad\qquad\qquad\qquad\qquad\qquad\qquad\varPsi_{q}(z)\quad\text{q-Polygamma Function} $$

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We have: $$ S = \sum_{n\geq 1}\frac{3^{n+1}}{6^n+1}=\sum_{m\geq 1}(-1)^{m+1} \sum_{n\geq 1}\frac{3^{n+1}}{6^{nm}}=9\sum_{m\geq 1}\frac{(-1)^{m+1}}{6^m-3} \tag{1}$$ and the last series is simple to accelerate: $$\begin{eqnarray*} S &=& 3-9\sum_{m\geq 2}\frac{(-1)^m}{6^m-3} = 3-9\sum_{m\geq 2}\frac{(-1)^m}{6^m}-27\sum_{m\geq 2}\frac{(-1)^m}{6^m(6^m-3)}\\&=&\color{blue}{\frac{851}{308}}-27\sum_{m\geq 3}\frac{(-1)^m}{6^m(6^m-3)}\tag{2}\end{eqnarray*}$$ Since the last series is a fast-converging series with alternating signs, $\left|S-\frac{851}{308}\right|\leq 5\cdot 10^{-4}$, and we may improve such approximation by just following the same approach. However, I do not think the original series has a nice "closed form".

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We have the identity $$ \begin{align} \sum_{k=1}^\infty\frac1{n^k\left(6^k+1\right)} &=\frac1{6n-1}-\sum_{k=1}^\infty\left(\frac1{(6n)^k}-\frac1{n^k\left(6^k+1\right)}\right)\\ &=\frac1{6n-1}-\sum_{k=1}^\infty\frac1{(6n)^k\left(6^k+1\right)}\tag{1} \end{align} $$ and we can bound the error in $(1)$ after $m$ terms by $$ \begin{align} \sum_{k=m+1}^\infty\frac1{(6n)^k\left(6^k+1\right)} &\le\sum_{k=m+1}^\infty\frac1{(36n)^k}\\ &=\frac1{(36n-1)(36n)^m}\tag{2} \end{align} $$


Using $(1)$, we can reduce the sum to converge as fast as we wish. For example, $$ \sum_{k=1}^\infty\frac{3^{k+1}}{6^k+1}=3\left(1-\frac1{11}+\frac1{71}-\frac1{431}+\sum_{k=1}^\infty\frac1{432^k\left(6^k+1\right)}\right)\tag{3} $$ and the error in $(3)$ after $m$ terms is less than $$ \frac3{2591\cdot2592^m}\tag{4} $$

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