Summation on floor function

Question

If $f(n)=\left[\sqrt{n}+\dfrac{1}{2}\right]$ when $n$ is a natural number and $[\cdot]$ represents the floor function, then find the value of

$$\sum_{n=1}^{\infty} \dfrac{2^{f(n)}+2^{-f(n)}}{2^n}$$.

How are we supposed to calculate this because when we run summation $2^n$ in denominator will change at each step but numerator will not change for each value of $n$? How to get started?

Answer 1

$\textbf{Note :} $ if $ k\in\mathbb{N}^{*} $, then $ \left(\forall p\in\left[\!\!\left[k^{2}-k+1,k^{2}+k\right]\!\!\right]\right),\ f\left(p\right)=k$.

\begin{aligned}\sum_{n=1}^{+\infty}{\frac{2^{f\left(n\right)}+2^{-f\left(n\right)}}{2^{n}}}&=\sum_{k=1}^{+\infty}{\left(\sum_{n=1}^{\left(k+1\right)^{2}-\left(k+1\right)}{\frac{2^{f\left(n\right)}+2^{-f\left(n\right)}}{2^{n}}}-\sum_{n=1}^{k^{2}-k}{\frac{2^{f\left(n\right)}+2^{-f\left(n\right)}}{2^{n}}}\right)}\\ &=\sum_{k=1}^{+\infty}{\sum_{n=k^{2}-k+1}^{k^{2}+k}{\frac{2^{f\left(n\right)}+2^{-f\left(n\right)}}{2^{n}}}}\\ &=\sum_{k=1}^{+\infty}{\left(2^{k}+2^{-k}\right)\sum_{n=k^{2}-k+1}^{k^{2}+k}{\frac{1}{2^{n}}}}\\ &=\sum_{k=1}^{+\infty}{2^{-k\left(k+1\right)}\left(2^{k}+2^{-k}\right)\left(2^{2k}-1\right)}\\ &=\sum_{k=1}^{+\infty}{\left(2^{2k-k^{2}}-2^{-2k-k^{2}}\right)}\\ &=\sum_{k=1}^{+\infty}{\left(2^{-k\left(k-2\right)}-2^{-\left(k+1\right)\left(k-1\right)}\right)}+\sum_{k=1}^{+\infty}{\left(2^{-\left(k+1\right)\left(k-1\right)}-2^{-\left(k+2\right)k}\right)}\\&=2^{-1\times\left(1-2\right)}+2^{-\left(1+1\right)\times\left(1-1\right)}\\ \sum_{n=1}^{+\infty}{\frac{2^{f\left(n\right)}+2^{-f\left(n\right)}}{2^{n}}}&=3\end{aligned}

$\textbf{Edit :}$ In the first line, we've use the fact that $ u_{\infty}-u_{0}=\sum\limits_{k=1}^{+\infty}{\left(u_{k+1}-u_{k}\right)} $, for any convergent sequence $ \left(u_{k}\right)_{k\in\mathbb{N}} $. In our case, $ u_{k}=\sum\limits_{n=1}^{k^{2}-k}{\frac{2^{f\left(n\right)}+2^{-f\left(n\right)}}{2^{n}}} $.

We've used it (Telescopic series) also in the $7^{\text{th}} $ line.

Answer 2

The visual access is best suited to gain a starting moment.

The following plot shows ticks of the sum not the series from 100 to 1000 in steps of 100:

This looks pretty constant.

So this converges really fast.

Only the first summand of the numerator, nominator contributes. The second one is very small pretty fast.

The prove is with the criteria of the convergence more rapid than the geometric series for both summands. The quotient of the series terms is already in the form of need

$\lim_{n\rightarrow\infty}\frac{a_n}{b_n}$

quotient criteria to the geometric series that is well known to be convergent. The above limit is zero since $n$ grows much faster than $\sqrt{n}$. The ceiling function does not matter for convergence.

Evaluated numerically the infinite sums or series value is $3$.

This can be interpreted as a sum of two convergent series:

2.56447 + 0.435532 = 3

So the second term can be rewritten as 3 - the first term. These compensate very early to give constant 3.

There are more identities for the series.