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For any Positive integer $n\;,$ Let $t(n)$ denote the integer closest to $\sqrt{n}\;,$

Then value of $\displaystyle \sum^{\infty}_{n=1}\frac{2^{t(n)}+2^{-t(n)}}{2^n}$

$\bf{My\; Try::}$ Here What i Understand is that $t(2)=1$ and $t(3) = 3$

So Let $t(n) = k\;,$ Then $\displaystyle \left(k-\frac{1}{2}\right)\leq \sqrt{n}<\left(k+\frac{1}{2}\right)\;,$ Where $k\in \mathbb{Z}$

So we get $$k^2-k+\frac{1}{4}\leq n<k^2+k+\frac{1}{4}$$

Now How can I solve after that, Help Required, Thanks

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juantheron
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1 Answers1

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$$\begin{eqnarray*}\sum_{n\geq 1}\frac{2^{t(n)}+2^{-t(n)}}{2^n}&=&\sum_{k\geq 1}(2^k+2^{-k})\sum_{n=k^2-k+1}^{k^2+k}\frac{1}{2^n}\\&=&\sum_{k\geq 1}(2^k+2^{-k})\frac{2^{2k}-1}{2^{k+k^2}}\\&=&\sum_{k\geq 1}\frac{2^{4k}-1}{2^{k^2+2k}}\end{eqnarray*}$$ but: $$ \frac{1}{2^{(k-1)^2}}-\frac{1}{2^{(k+1)^2}}=\frac{2^{4k}-1}{2^{k^2+2k+1}} $$ hence the original series is a telescopic series.

Jack D'Aurizio
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  • Thanks Jack D'Aurizio, But i did not understand why we take Inner summation from $n=k^2-k+1$ to $k^2+k;,$ plz explain me, Thanks – juantheron Jun 07 '16 at 04:36
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    @juantheron: your bound $$ k^2-k+\frac{1}{4}\leq n \leq k^2+k+\frac{1}{4}$$ is right, but $n$ has to be an integer, hence we may simply state $k^2-k+1\leq n\leq k^2+k$. – Jack D'Aurizio Jun 07 '16 at 04:37
  • Thanks Jack I understand Right side inequality $\displaystyle n\leq k^2+k < k^2+k+\frac{1}{4};,$ But still i did not understand left side Inequality. – juantheron Jun 07 '16 at 04:40
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    @juantheron: they are the same thing. If $n\geq k^2-k+\frac{1}{4}$ and $n$ is an integer, $n\geq k^2-k+1$. – Jack D'Aurizio Jun 07 '16 at 04:41