Calculate the converging limit of this series

Question

$$ \left\{ \begin{split} S(0) &= 1\\ S(i+1) &= \left(1+\frac{1}{2i} \right) S(i)\qquad \text{for}~ i > 0 \end{split} \right. $$ I somehow figured out that this series must converge, but could not find the value where it converges.

Answer 1

Assuming you mean $S(1) = 1$ - otherwise your sequence is still not well defined - the sequence does not converge:

• Set $l_i := \ln S(i)$ for $i \in \mathbb{N}$
• $\Rightarrow l_{i+1} = \ln(2i+1) - \ln (2i) + l_i$ and $l_1 =0$
• $\Rightarrow l_{i+1} = l_1 + \sum_{k=1}^i(l_{i+1} - l_i) = \sum_{k=1}^i(\ln(2i+1) - \ln (2i))$
• The mean value theorem gives $\ln(2i+1) - \ln (2i)> \frac{1}{2i+1}$

$$\Rightarrow l_{i+1} > \sum_{k=1}^i\frac{1}{2i+1}> \sum_{k=1}^i\frac{1}{2i+2}=\frac{1}{2}\sum_{k=1}^i\frac{1}{i+1}\stackrel{i \to \infty}{\longrightarrow} \infty$$

Hence,

$$S(i) = e^{l_i} \stackrel{i \to \infty}{\longrightarrow} \infty$$

Answer 2

If it was $$\left\{ \begin{split} S_1 &= 1\\ S_{i+1} &= \left(1+\frac{1}{2i} \right) S_i\qquad \text{for}~ i > 1 \end{split} \right.$$ using Pochhammer symbols, the result would be $$S_i=\frac{\left(\frac{3}{2}\right)_{i-1}}{(1)_{i-1}}$$ Expanded as series for large values of $i$ we should have $$S_i=\frac{2 \sqrt{i}}{\sqrt{\pi }}-\frac{1}{4 \sqrt{\pi i }}+\frac{1}{64 \sqrt{\pi i^3 }}+O\left(\frac{1}{i^2}\right)$$ which does not converge.

Trying for $i=10$, the exact result is $$S_{10}=\frac{230945}{65536}\approx 3.52394$$ while the above truncated series would give $\frac{12641}{640 \sqrt{10 \pi }}\approx 3.52392$