jmc

We hope this sum telescopes. We really hope this sum telescopes.

One thing to think about is what happens when we add up the first few terms. (The sum of the first few terms of an infinite series is called a partial sum.) For example, when we add the first three terms of the series, we obtain a fraction whose denominator is $$(5+1)(5^2+1)(5^4+1).$$We can make this product nicely collapse by multiplying it by $5-1$: $$\begin{array}{rl}(5-1)(5+1)(5^2+1)(5^4+1)& =(5^2-1)(5^2+1)(5^4+1)\\ & =(5^4-1)(5^4+1)\\ & =5^8-1.\end{array}$$More generally, if we add the first $n$ terms of the series, we can obtain a fraction with denominator $5^{2^n}-1.$ The next term in the series has a denominator of $5^{2^n}+1.$ Since we want the sum to telescope, we can consider the difference $$\frac{1}{5^{2^n}+1}-\frac{1}{5^{2^n}-1}=\frac{2}{5^{2^{n+1}}-1}.$$Multiplying both sides by $2^n,$ we get $$\frac{2^n}{5^{2^n}+1}-\frac{2^n}{5^{2^n}-1}=\frac{2^{n+1}}{5^{2^{n+1}}-1}.$$Thus, $$\frac{2^n}{5^{2^n}+1}=\frac{2^n}{5^{2^n}-1}-\frac{2^{n+1}}{5^{2^{n+1}}-1}.$$The given series then telescopes as follows: $$\begin{array}{rl}\frac{1}{5+1}+\frac{2}{5^2+1}+\frac{4}{5^4+1}+\cdots & =\left(\frac{1}{5-1}-\frac{2}{5^2-1}\right)+\left(\frac{2}{5^2-1}-\frac{4}{5^4-1}\right)+\left(\frac{4}{5^4-1}-\frac{8}{5^8-1}\right)+\cdots \\ & =\boxed{\frac{1}{4}}.\end{array}$$