Does this weighted sum of reciprocals of Fermat numbers,
$$ F=\sum_{k=0}^{\infty}\dfrac{2^{k}}{2^{2^{k}}+1} $$
have a nice closed form? Wolfram says it's $1$.
Thanks.
Does this weighted sum of reciprocals of Fermat numbers, $$ F=\sum_{k=0}^{\infty}\dfrac{2^{k}}{2^{2^{k}}+1} $$ have a nice closed form? Wolfram says it's $1$. Thanks. |
||||
show 2 more comments |
Hint: Try to guess and prove a formula for partial sums $$ S(n)=\sum_{k=0}^n\frac{2^k}{2^{2^k}+1}. $$ Here $$ S(0)=\frac13,\ S(1)=\frac{11}{15},\ S(2)=\frac{247}{255},\ldots $$ See a pattern for $1-S(n)$? |
|||||||||||||
|
This might be another way of looking at Jyrki's hint, but here is the way I did this: $$ \begin{align} \color{#C0C0C0}{1-\frac1{2-1}+}\frac1{2+1}&=1-\frac2{4-1}\\ 1-\frac{2}{4-1}+\frac{2}{4+1}&=1-\frac{4}{16-1}\\ 1-\frac{4}{16-1}+\frac{4}{16+1}&=1-\frac8{256-1}\\ 1-\frac8{256-1}+\frac8{256+1}&=1-\frac{16}{65536-1}\\ &\vdots \end{align} $$ |
|||||||||||||||||
|
NSum
only does numerical sums. IfSum
cannot find a closed form, it callsNSum
to compute numerically. Therefore, if you are only looking for a numerical approximation, thenNSum
will always be faster. – robjohn♦ Apr 8 at 14:29