What is the limit of this divergent infinite product multiplied by an exponential?

Solution 1:

Define $y_k = \frac{1}{c^k} \prod_{i=1}^k(1 + e^{b/c^i})$.

Claim 1: If $c>2$, then $\lim_{k\rightarrow\infty} y_k=0$. If $1\leq c<2$, then $\lim_{k\rightarrow\infty} y_k = \infty$.

Proof for case $c>2$: Assume $c = 2 + \delta$ for some $\delta>0$. There exists a value $k^*$ such that $e^{b/c^i} \leq 1 + \delta/2$ for all $i \geq k^*$. Then for all $k>k^*$ we have: \begin{align} 0 \leq y_k &= \frac{1}{c^k} \prod_{i=1}^k (1 + e^{b/c^i})\\ &= \left(\frac{1}{c^{k^*}}\prod_{i=1}^{k^*}(1+e^{b/c^i})\right)\frac{1}{c^{k-k^*}}\prod_{i=k^*+1}^{k} (1+e^{b/c^i})\\ &\leq \left(\frac{1}{c^{k^*}}\prod_{i=1}^{k^*}(1+e^{b/c^i})\right)\frac{(2+\delta/2)^{k-k^*}}{c^{k-k^*}}\\ &= \left(\frac{1}{c^{k^*}}\prod_{i=1}^{k^*}(1+e^{b/c^i})\right)\left(\frac{2+\delta/2}{2 + \delta}\right)^{k-k^*} \rightarrow 0\\ \end{align} So $\lim_{k\rightarrow\infty} y_k = 0$.

Proof for case $1 < c < 2$: Similar.

Proof for case $c=1$: $y_k = (1+e^{b/c})^k \rightarrow \infty$.

Claim 2: If $0 < c < 1$ and $b\geq 0$ then $\lim_{k\rightarrow\infty} y_k = \infty$.

Proof: We know $1+e^{b/c^i} \geq 1$ for all $i \geq 1$, and so $y_k \geq 1/c^k\rightarrow\infty$.

Claim 3: If $c=2$ and $b\geq 0$ then $1 \leq \lim_{k\rightarrow\infty} y_k \leq e^b$.

Proof: Since $b \geq 0$ we know $1 \leq e^{b/2^i}$ for all $i$, and hence:

\begin{align} y_k &= \frac{1}{2^k} \prod_{i=1}^k(1 + e^{b/2^i}) \\ &\leq \frac{1}{2^k} \prod_{i=1}^k (e^{b/2^i} + e^{b/2^i}) \\ &= \prod_{i=1}^k e^{b/2^i} \end{align}

Thus: \begin{align} \log(y_k) &\leq \sum_{i=1}^k \frac{b}{2^i} \rightarrow b \end{align} and so $\lim_{k\rightarrow\infty} y_k \leq e^b$.

For an exact answer for $c=2, b\neq 0$, why not just apply your same method?

\begin{align} \int_0^1 e^{bx} dx &= \int_0^{1/2} e^{bx}dx + \int_0^{1/2} e^{b(x+1/2)}dx \\ &= (1+e^{b/2})\int_0^{1/2} e^{bx}dx \\ &= (1+e^{b/2})\left[ \int_0^{1/4} e^{bx}dx + \int_0^{1/4} e^{b(x+1/4)}dx\right]\\ &= (1+e^{b/2})(1+e^{b/4})\int_0^{1/4}e^{bx}dx \end{align}

and so on, with the left-hand-side giving the answer of $\frac{e^b-1}{b}$? So if $c=2$ and $b\neq 0$ we get: $$ \lim_{k\rightarrow\infty} y_k = \frac{e^b-1}{b} $$

Notice that this is consistent with Claim 3, since $1 < \frac{e^b-1}{b} < e^b$ whenever $b>0$.