To find the sum of a finite series : $\sum_{i=1}^n \frac{2}{3^i-1}$

Question

I need to prove that : $\sum_{i=1}^n \frac{2}{3^i-1}$ < $ \frac{3}{2} $

Further if possible can you tell me how we can find the sum of a finite series like that.

Answer 1

$$\begin{eqnarray*}\color{red}{S}=\sum_{n\geq 1}\frac{2}{3^n-1}=2\sum_{n\geq 1}\frac{3^{-n}}{1-3^{-n}}=2\sum_{k\geq 1}\sum_{n\geq 1}3^{-kn}&=&2\sum_{h\geq 1}d(h)\,3^{-h}\\&\color{red}{<}&2\sum_{h\geq 1}h\,3^{-h}=2\cdot\frac{3}{4}=\color{red}{\frac{3}{2}}.\end{eqnarray*}$$ Such series (Lambert-like) tend not to have a nice closed form, but their numerical computation is simple since we may design on-purpose acceleration techniques, based on the fact that the main term is approximately geometric - see this similar question, for instance.

In the current case we have $$ S = 1+\sum_{n\geq 1}\frac{2}{3^{n+1}-1} = 1+\frac{1}{3}+\sum_{n\geq 1}\frac{2}{9^{n+1}-3^{n+1}}=\frac{49}{36}+\sum_{n\geq 1}\frac{2}{9^{n+2}-3^{n+2}} $$ so we may improve the previous upper bound in the following way: $$ S< \frac{49}{36}+\sum_{n\geq 1}\frac{2}{9^{n+1}\left(1-\frac{1}{27}\right)}=\frac{49}{36}+\frac{1}{312}=\frac{1277}{936}$$ where the RHS agrees with the exact value of $S$ up to four significative figures.
Actually $S\approx 1.364307$.

Answer 2

Here is a pretty quick solution, for $i>1$ we have

$$\frac{2}{3^i-1} <\frac{2}{3^i - 3^{i-1}}=\frac{1}{3^{i-1}},$$

so for $n>1$ we have $$\sum_{i=1}^n \frac{2}{3^i-1}<\sum_{i=0}^\infty \frac{1}{3^i}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}.$$ For $n=1$ the inequality is trivial, so we have proven the claim.