Convergence of $P=\prod_{k=1}^{\infty}\left(1+\frac{1}{5^k}\right) $

Question

Convergence of $$P=\prod_{k=1}^{\infty}\left(1+\frac{1}{5^k}\right) $$

Will this product converges to finite limit?

My try:

we have $$P=1+\frac{1}{5}+\frac{1}{5^2}+2\frac{1}{5^3}+\cdots+2\frac{1}{5^7}+\cdots\infty$$

Answer 1

A possible fast way is to note that

$$\prod (1+a_k)\leq\exp \left(\sum a_k\right)$$ and $$\sum a_k =\sum 5^{-k}\leq\sum k^{-2}<\infty.$$

Of course this approach works for all infinite products of the form $\prod(1+\frac1{a^k})$ as long as $a>1$

Answer 2

Using $\ln x\leq x-1$,

$$\ln P=\sum_{k=1}^\infty\ln\left(1+\frac1{5^k}\right)\leq\sum_{k=1}^\infty\left(1+\frac1{5^k}-1\right)=\frac14$$

As $\ln P\approx 0.23$, the bound is quite tight.

Answer 3

Not recommended, but invoking Bertrand's postulate allows one to avoid taking logs:

$$1+5^{-k}\lt{1\over1-5^{-k}}\lt{1\over1-2^{-2k}}\lt{1\over1-p_k^{-2}}$$

so

$$\prod_{k=1}^\infty(1+5^{-k})\lt\prod_p{1\over1-p^{-2}}=\zeta(2)=\sum_{n=1}^\infty{1\over n^2}\lt\infty$$