Determine the closed form for

$$\sum_{n=1}^{\infty} \frac{1}{2^n \left ( 1+ \sqrt[2^n]{2}\right)}$$

There were many high-rep users on M.SE asking how one could even assume such a closed-form exists. A quick glance at this shows that the sum should be bounded from above by $1/2$. Otherwise, anything goes, right?

Actually, this is a problem that is made more difficult than it really is because of the horrible notation. Hope springs when the above sum is rewritten:

$$\sum_{n=1}^{\infty} \frac{2^{-n}}{ 1+ 2^{2^{-n}}}$$

Now this looks like something we can work with. What I have learned over the years is to look for the anti-difference of the summand, so we can get a telescoping sum and easily compute the original sum. In this case, it is fortuitous that

$$\frac{2^{-n}}{2^{2^{-n}}-1}-\frac{2^{-(n-1)}}{2^{2^{-(n-1)}}-1} = \frac{2^{-n}}{2^{2^{-n}}+1} $$

Thus we indeed have a telescoping sum. However, note that

$$\lim_{n \to \infty} \frac{2^{-n}}{2^{2^{-n}}-1} = \frac{1}{\log{2}}$$

Therefore the sum is

$$a_1-a_0 + a_2-a_1 + a_3-a_2 + \ldots + \frac{1}{\log{2}} = \frac{1}{\log{2}} – a_0$$

where

$$a_n = \frac{1}{2^n \left ( 2^{2^{-n}}-1\right)}$$

or

$$\sum_{n=1}^{\infty} \frac{1}{2^n \left ( 1+ \sqrt[2^n]{2}\right)}= \frac{1}{\log{2}} – 1 \approx 0.442695$$

Hello from Russa, Ron

How about such a series with n-th term

(3^(-n))*(3^(3^(-n)-2)/(3^(2*3^(-n))+3^(3^(-n))+1)

and the sum of S=1/ln(3)-1/2 ?

One can see, that integer 3 may be replaced by any positive interger k>1. Then we have the sum

S=1/ln(k)-1/(k-1).

Certainly you need replace the n-th term. Hint, you must use two beautiful formulas

a^k-1=(a-1)(a^(k-1)+a^(k-2)+…+a+1),

a^k+a^(k-1)+…+a^2+a-k=(a-1)(a^(k-1)+2*a^(k-2)+3*a^(k-3)+…+(k-1)*a-k)

I hope you know how. If not, email me and I reply to you