Dangerous sigma

For any positive integer $n$ , let $n'$ denote the closest integer to $\sqrt n$ , then $\displaystyle \sum_{n=1}^\infty \frac{2^{n'} + \left(\frac12 \right)^{n'}}{2^n} = \ ?$

I don't seem to understand how to even approach this question, any help would be appreciated.

Edit: The question has been turned into a problem by @Chew-Seong Cheong sir. It can be accessed here.

#Calculus

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Comments

Using Wolfram Alpha the answer is 3. See here.

Chew-Seong Cheong - 1 year, 1 month ago

An algebraic answer would have been more helpful sir

Adhiraj Dutta - 1 year, 1 month ago

I would not want to try. No algebraic method to handle nearest integers $\lfloor x + \frac 12 \rfloor$ .

Chew-Seong Cheong - 1 year, 1 month ago

OK. I will try if I got I will set it as a question. I have some idea now.

Chew-Seong Cheong - 1 year, 1 month ago

Check out the solution here.

Chew-Seong Cheong - 1 year, 1 month ago

@Chew-Seong Cheong – Thank you sir! Very grateful.

Adhiraj Dutta - 1 year, 1 month ago

@Chew-Seong Cheong – Did you delete the solution to my recent problem? @Chew-Seong Cheong

Adhiraj Dutta - 11 months ago

@Adhiraj Dutta – Yes, I don't think my solution was correct.

Chew-Seong Cheong - 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$