Does $\sum^{\infty}_{n=1} \frac{1}{a_n+1}$ with $a_{n+1}=a_n(a_n+1)$ and $a_1=\frac12$ have a closed form?

Question

Mr Beast here. I've been working on a complicated math problem, and I obtained a recursive sequence $a_{n+1}=a_n(a_n+1)$ and that $a_1=\frac 12$. I've been thinking, is it possible to convert this to closed form, or otherwise evaluate $\sum^{\infty}_{n=1} \frac{1}{a_n+1}$?

The original question said that $a_{1}=\frac{1}{2}$ and that $a_{n+1}=a_{n}\left(a_{n}+1\right),$ asked me to prove that $\frac{1}{a_{1}+1}+\frac{1}{a_{2}+1}+\frac{1}{a_{3}+1}+\cdots+\frac{1}{a_{n}+1}<2$ for all $n$ larger than $1 .$ I wondered if it was possible to convert into closed form to evaluate the sum definitely. I originally thought to solve it through a telescoping series, but the denominator leaves me unsure of what to do.

Answer 1

This is not an answer, but deals with the underlying problem disclosed by OP in a comment to the question.

Consider the terms $$b_i = \frac{1}{a_i + 1}$$where $a_1 = \frac{1}{2}$ and $a_{i+1} = a_i^2 + a_i$: $$b_1 = \frac{2}{3}, \, b_2 = \frac{4}{7}, \, b_3 = \frac{16}{37}, \, b_4 = \frac{256}{1033}, \, b_5 = \frac{65536}{868177}, \, \dots$$ Your task is to find a sequence that does have a closed form expression, for example $$\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6} \approx 1.64493407$$ where each term $c_k = \frac{1}{k^2}$. The sequence must be such that $$c_k \ge b_k \quad \text{for all} \quad k \ge n$$
Your proof is then based on sums $$\begin{aligned} B &= \sum_{k=1}^\infty b_k = B_1 + B_2, \quad B_1 = \sum_{k=1}^{n-1} b_k, \quad B_2 = \sum_{k=n}^\infty b_k \\ C &= \sum_{k=1}^\infty c_k = C_1 + C_2, \quad C_1 = \sum_{k=1}^{n-1} c_k, \quad C_2 = \sum_{k=n}^\infty c_k = C - C_1 \\ \end{aligned}$$ where $B$ and $B_2$ are not known exactly, only $B_1$ is known exactly; but $C$, $C_1$, and $C_2$ are known exactly.

Because $$c_k \ge b_k \quad \text{ for } \quad k \ge n, \quad C_2 \ge B_2$$ and if you prove that $$B_1 + C_2 \lt 2$$ (which is something you have closed expressions for, if the infinite sum $C$ has a known closed expression), you have proven that $$B_1 + B_2 = B \lt 2$$

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The series to focus on those related to the Riemann Zeta function at even values: $$\zeta(n) = \sum_{k=1}^\infty \frac{1}{k^n}$$ The particular one I was thinking of is $$\zeta(20) = \sum_{k=1}^\infty \frac{1}{k^{20}} = \frac{174611 \, \pi^{20}}{1531329465290625} \approx 1.000000953962$$ with $n = 10$.

You see, $$c_k = \frac{1}{k^{20}} \ge b_k \quad \text{for all} \quad k \ge 10$$ In particular, $$\begin{aligned} b_{8} \approx 1.4255 \cdot 10^{-9} \\ b_{9} \approx 2.0321 \cdot 10^{-18} \\ b_{10} \approx 4.1292 \cdot 10^{-36} \\ b_{11} \approx 1.7051 \cdot 10^{-71} \\ c_{8} \approx 8.6736 \cdot 10^{-19} \lt b_{8} \\ c_{9} \approx 8.2253 \cdot 10^{-20} \lt b_{9} \\ c_{10} = 1.0000 \cdot 10^{-20} \gt b_{10} \\ c_{11} \approx 1.4864 \cdot 10^{-21} \gt b_{11} \\ \end{aligned}$$ so we know that $$\sum_{k=10}^\infty b_k \lt \sum_{k=10}^\infty \frac{1}{k^{20}}$$ and because $$\sum_{k=10}^\infty \frac{1}{k^{20}} = \frac{174611 \, \pi^{20}}{1531329465290625} - \frac{106662370149283833915501942267857492605696794711249007378798472708401}{106662268397529335923611516553737065783558248857600000000000000000000}$$ we can now calculate the exact sum, $$\left( \sum_{k=1}^{9} b_k \right) + \left( \sum_{k=10}^\infty \frac{1}{k^{20}} \right) \lt 2$$ and therefore $$\sum_{k=1}^\infty b_k \lt 2$$

Because floating-point numerical calculations do not have enough precision here, I used $$C = \sum_{k=1}^\infty \frac{1}{k^{20}} = \frac{174611 \, \pi^{20}}{1531329465290625}$$ and $$\begin{aligned} C_1 &= \sum_{k=1}^9 \frac{1}{k^{20}} \\ ~ &= \frac{106662370149283833915501942267857492605696794711249007378798472708401}{106662268397529335923611516553737065783558248857600000000000000000000} \\ \end{aligned}$$ and $$\begin{aligned} B_1 &= \sum_{k=1}^9 b_k \\ ~ &= \frac{6494086337598191931522860064611183626609948977242275075491600909077850578669643051565257331009366963694648505112737011786954151363577920417599029974920037685716132002271729535802246547158018}{3247043168799095965761430032305591820008878453592436087532812953641848353074504436078825354366464203708206267593141994294414224765240674053815444080703044269735007407109149741117947776621057} \\ \end{aligned}$$ and verified that $B_1 + C - C_1 \lt 2$ via $$C + B_1 - C_1 - 2 \lt 0$$ The left side, for evaluation in e.g. Maxima (a free CAS program, wxMaxima being its graphical version), is

174611*(%pi^20) / 1531329465290625 - 16492253350531080128239160104845001622128429492012917644483996917752120075662103069882162351389153630866311951764035014010230695125685435114364416532374072620244556261880185006228929268661858194735280962676368978555791749162077866217824355140382739573209517 / 16492237617562539365036586070920801101220100452147363781860837846145943893309011792682018865262830776584392846863644670314594880250899416573838450710644960492510494335952071191890802960534017689711520151393881057079583788816561717064499200000000000000000000

and Maxima gives it an approximate value $-6.6613\cdot10^{-16}$, which indeed is less than zero, proving the series sum $\sum_{k=1}^\infty b_k \lt 2$.

Answer 2

Hint: show (using induction on $m$) that $$\sum_{n=1}^m\frac{1}{a_n+1}=\frac1{a_1}-\frac1{a_{m+1}}.$$

Answer 3

Note that $$ \frac1{a_{n+1} } = \frac1{a_n(a_n+1)}=\frac{(a_n+1)-a_n}{a_n(a_n+1)} = \frac1{a_n} - \frac1{a_n+1}$$ Therefore $$ \frac1{a_n+1}=\frac1{a_n}-\frac1{a_{n+1}}$$ Summing in $n$ gives a telescoping series, $$ \sum_{k=1}^n\frac1{a_k+1} = \sum_{k=1}^n\frac1{a_k}-\frac1{a_{k+1}} = \frac1{a_1} - \frac1{a_{n+1}}.$$ Plugging in $a_1=1/2$, we see that $$ \sum_{k=1}^n\frac1{a_k+1} = 2 - \frac1{a_{n+1}}.$$

Note that $a_1=1/2,\ a_2=3/4,\ a_3=21/16>1$, and from that point on the equation gives $$ a_{n+1} = a_n(a_n+1) \ge a_n+1$$ so that $a_n\to\infty$. Hence $\frac1{a_{n+1}}\to 0$ as $n\to\infty$, which means that $$ \fbox{$\sum_{k=1}^\infty\frac1{a_k+1} = 2. $}$$

In particular, since $\frac1{a_n+1}>0$ for all $n$, this proves (as is the 'original question' in the comments) that the finite sum of only the first $n$ terms is strictly smaller than $2$ for all $n$.