Computing the exact value of $\sum_{n=1}^\infty \left(\frac{2n+3}{3n+2}\right)^n$ I found this problem in my textbook, and I know that it converges, but I wanted to know if there was a way to find the exact value of the convergence (similar to what Euler did with the sum of reciprocal squares).
I tried to rewrite the sum as a power series of sorts, but I don't know if it's correct, or if it made anything more complicated. 
Steps:
$$\lim\limits_{a \to \infty} \sum_{n=1}^a \Big(\frac{2n+3}{3n+2}\Big)^n=\lim\limits_{a \to \infty} \sum_{n=1}^a \Big(\frac{2}{3} + \frac{5/3}{3n+2} \Big)^n =\lim\limits_{a \to \infty} \sum_{n=1}^a \Big(\frac{1}{3}\Big)^n\Big(2+\frac{5}{3n+2}\Big)^n$$
$$=\lim\limits_{a \to \infty} \sum_{n=1}^a \Big(\frac{1}{3}\Big)^n \sum_{m=0}^n 2^{n-m}\Big(\frac{5}{3n+2}\Big)^m \binom{n}{m}= \lim\limits_{a \to \infty} \sum_{n=1}^a \Big(\frac{2}{3}\Big)^n \sum_{m=0}^n \Big(\frac{5}{2(3n+2)}\Big)^m \binom{n}{m}$$
$$=\lim\limits_{a \to \infty}\sum_{n=1}^a \Big(\frac{2}{3}\Big)^n \sum_{m=0}^n\Big(\frac{5}{6n+4}\Big)^m \binom{n}{m}$$
I hit a wall here because I am not sure what do with the double sum part of the problem.
(Note: I made the top limit of the outer sum $a$, and took the limit as $\,$$a\to\infty$$\,$ because when I tried to make a table to evaluate the double sum, I wanted to use something finite in order to get a finite answer.)
EDIT: Is there an explicit formula if $a$ does not approach $\infty$?
 A: This series is closely related to the well-known identities of Bernoulli which have no closed form and so might go some way in explaining why obtaining one would be difficult  (although...it's pretty obvious).  To see this consider a more general series
$$S=\sum\limits_{n=1}^{\infty }{{{\left( \frac{{{a}_{1}}n+{{a}_{2}}}{{{b}_{1}}n+{{b}_{2}}} \right)}^{n}}}=\sum\limits_{n=1}^{\infty }{{{\left( \frac{{{a}_{1}}}{{{b}_{1}}} \right)}^{n}}{{\left( 1+\frac{{{r}_{a}}-{{r}_{b}}}{n+{{r}_{b}}} \right)}^{n}}}=\sum\limits_{n=1}^{\infty }{{{R}^{n}}\sum\limits_{m=0}^{n}{\left( \begin{matrix}
   n  \\
   m  \\
\end{matrix} \right)\frac{{{\left( {{r}_{a}}-{{r}_{b}} \right)}^{m}}}{{{\left( n+{{r}_{b}} \right)}^{m}}}}}\,\,$$
where the ratios of coefficients are defined ${{r}_{a}}={{a}_{2}}/{{a}_{1}}$, ${{r}_{b}}={{b}_{2}}/{{b}_{1}}$, $R={{a}_{1}}/{{b}_{1}}$, and the coefficients are chosen to ensure convergence.  Taking out the first term in the second summation and using
$$\frac{1}{{{\left( n+r \right)}^{m}}}=\frac{1}{\Gamma \left( m \right)}\int\limits_{0}^{\infty }{{{e}^{-u\left( n+r \right)}}{{u}^{m-1}}du}$$
yields
$$S=\frac{R}{1-R}+\sum\limits_{n=1}^{\infty }{{{R}^{n}}\sum\limits_{m=1}^{n}{\left( \begin{matrix}
   n  \\
   m  \\
\end{matrix} \right){{\left( {{r}_{a}}-{{r}_{b}} \right)}^{m}}}}\frac{1}{\Gamma \left( m \right)}\int\limits_{0}^{\infty }{{{e}^{-u\left( n+{{r}_{b}} \right)}}{{u}^{m-1}}du}$$
which can be written as
$$S=\frac{R}{1-R}+\sum\limits_{n=1}^{\infty }{{{R}^{n}}\left( {{r}_{a}}-{{r}_{b}} \right)\sum\limits_{m=0}^{n-1}{\left( \begin{matrix}
   n  \\
   n-m-1  \\
\end{matrix} \right){{\left( {{r}_{a}}-{{r}_{b}} \right)}^{m}}}}\frac{1}{m!}\int\limits_{0}^{\infty }{{{e}^{-u\left( n+{{r}_{b}} \right)}}{{u}^{m}}du}$$
Note the Laguerre polynomial
$$L_{n-1}^{1}\left( x \right)=\sum\limits_{m=0}^{n-1}{\frac{1}{m!}\left( \begin{matrix}
   n  \\
   n-m-1  \\
\end{matrix} \right){{\left( -x \right)}^{m}}}$$
and so
$$S=\frac{R}{1-R}+\left( {{r}_{a}}-{{r}_{b}} \right)\sum\limits_{n=1}^{\infty }{{{R}^{n}}}\int\limits_{0}^{\infty }{L_{n-1}^{1}\left( \left( {{r}_{b}}-{{r}_{a}} \right)u \right){{e}^{-u\left( n+{{r}_{b}} \right)}}du}$$
By the generating function
$$\sum\limits_{n=1}^{\infty }{{{t}^{n}}L_{n-1}^{\left( 1 \right)}\left( x \right)}=\frac{t{{e}^{-\frac{tx}{1-t}}}}{{{\left( 1-t \right)}^{2}}}$$
the series then reduces to
$$S=\frac{R}{1-R}+\left( {{r}_{a}}-{{r}_{b}} \right)\int\limits_{0}^{\infty }{\frac{{{\operatorname{Re}}^{-\left( 1+{{r}_{b}} \right)u}}{{e}^{-\frac{{{\operatorname{Re}}^{-u}}}{1-{{\operatorname{Re}}^{-u}}}\left( {{r}_{b}}-{{r}_{a}} \right)u}}}{{{\left( 1-{{\operatorname{Re}}^{-u}} \right)}^{2}}}du}$$
which after a change of variable and substitution for the ratios
$$\sum\limits_{n=1}^{\infty }{{{\left( \frac{{{a}_{1}}n+{{a}_{2}}}{{{b}_{1}}n+{{b}_{2}}} \right)}^{n}}}=\frac{{{a}_{1}}}{{{b}_{1}}-{{a}_{1}}}+\left( {{a}_{2}}{{b}_{1}}-{{a}_{1}}{{b}_{2}} \right)\int\limits_{0}^{1}{\frac{{{x}^{\frac{{{b}_{2}}-{{a}_{2}}x}{{{b}_{1}}-{{a}_{1}}x}}}}{{{\left( {{b}_{1}}-{{a}_{1}}x \right)}^{2}}}dx}$$
Perhaps we can call this a ‘sophomore’s reality’, which certainly incorporates a dream  $\left( {{a}_{1}},{{a}_{2}},{{b}_{1}},{{b}_{2}} \right)=\left( 0,\pm 1,1,0 \right)$
$$\sum\limits_{n=1}^{\infty }{\frac{1}{{{n}^{n}}}}=\int\limits_{0}^{1}{\frac{1}{{{x}^{x}}}dx}\\  \sum\limits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}}{{{n}^{n}}}}=\int\limits_{0}^{1}{{{x}^{x}}dx}$$
However the OP's series remains grounded and awake
$$\sum\limits_{n=1}^{\infty }{{{\left( \frac{2n+3}{3n+2} \right)}^{n}}}=2+5\int\limits_{0}^{1}{\frac{{{x}^{\frac{2-3x}{3-2x}}}}{{{\left( 3-2x \right)}^{2}}}dx}$$
