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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 $\infty$?

Rodrigo de Azevedo
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transcenmental
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    I would be very surprised if this had a closed form. But I have been surprised before. – Robert Israel May 05 '17 at 00:37
  • I don't know how to calculate the exact sum, but you can prove convergence by noting that for $n\geq 2$, $\frac{2n+3}{3n+2}\leq \frac{7}{8}\implies Sum\leq 1+\sum_{n=2}^\infty\left(\frac{7}{8}\right)^n = \frac{57}{8}$ – Lazy Lee May 05 '17 at 00:37
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    What makes you think this could have a closed sum? – Saketh Malyala May 05 '17 at 00:42
  • @SakethMalyala The ratio test of sequences shows that if $f_n=(\frac{2n+3}{3n+2})^n$, and if $\lvert f_{n+1}/f_n \rvert \lt 1$, then the infinite sum of $f_n$ converges. – transcenmental May 05 '17 at 00:53
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    That just proves convergence. I believe what @sakethmalyala was asking what made you think that there is a solution comprised of elementary functions. For example, it is known that the sum of reciprocal cubes converges, but there is no exact value/closed solution known. – infinitylord May 05 '17 at 01:06
  • @infinitylord oh... well that changes everything. – transcenmental May 05 '17 at 01:19
  • @infinitylord I also found it odd that this series, as $n$ approaches $\infty$, the ratio of consecutive terms, tends to something like 0.6, rather than 0.6 repeating (2/3), which is what intuitively one would expect this to come out to. – transcenmental May 05 '17 at 03:25
  • Often when one transforms a single summation into a double summation, a useful next step is to reverse the order of summation. I tried that here but didn't get anywhere. – Julian Rosen May 05 '17 at 14:11
  • If I've calculated right it's $\enspace\displaystyle \sum\limits_{n=1}^\infty \left(\frac{2n+3}{3n+2}\right)^n =$ $\displaystyle = -1 + \left(\frac{2}{3}\right)^{\frac{1}{3}} \frac{d}{dt} \left[ t^{-\frac{1}{3}} \int\limits_0^\infty \frac{\exp\left(-\frac{3}{2} W\left(-xe^{-\frac{3}{2}x/t}\right)\right)}{1+W\left(-xe^{-\frac{3}{2}x/t}\right)} e^{-\frac{2}{3}x/t}dx\right] |_{t=\frac{2}{3}} ,$ . $\enspace W(x)$ ist the Lambert W-Function. – user90369 Mar 02 '18 at 15:24
  • $$\sum_{n=1}^{\infty}\left(\frac{2n+3}{3n+2}\right)^n \approx \frac{1}{4}+\frac{\pi}{\ln 5 -C}$$ where $C$ is Cahen's constant: $$C=\frac{1}{2}+\frac{1}{7}+\frac{1}{1807}+\frac{1}{1060056950807}+...\approx 0.64341054629...$$ – Suhrid Saha Oct 24 '18 at 16:55

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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 $$\sum\limits_{n=1}^{\infty }{{{\left( \frac{2n+3}{3n+2} \right)}^{n}}}=2+5\int\limits_{0}^{1}{\frac{{{x}^{\frac{2-3x}{3-2x}}}}{{{\left( 3-2x \right)}^{2}}}dx}$$

mathstackuser12
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