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

that sum is not infinite and I don't know where you got it from — mathworker21, Oct 08 '18 at 17:33
$$\left(-1;\dfrac{1}{5}\right)\infty = \prod{k=0}^\infty\left(1 + \dfrac{1}{5^k}\right)$$ so $$\prod_{k=1}^\infty \left(1+\dfrac{1}{5^k}\right) = \dfrac{\left(-1;\dfrac{1}{5}\right)_\infty}{2}$$ Wolframalpha — SlipEternal, Oct 08 '18 at 17:43
Since $\sum_{k\geq 1}\frac{1}{5^k}$ is convergent your product is convergent. For accurate approximations see this answer. — Jack D'Aurizio, Oct 08 '18 at 18:15

score 2 · Answer 1 · answered Oct 08 '18 at 17:48

2

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

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

answered Oct 08 '18 at 17:48

Jam

score 1 · Answer 2 · answered Oct 08 '18 at 17:44

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 $a>1$

answered Oct 08 '18 at 17:44

b00n heT

score 1 · Answer 3 · answered Nov 12 '18 at 15:19

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

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

3 Answers3