Can you find the closed form expression for this series?

Since Brilliant users are really brilliant, I thought that it would be great to discuss with them my problem. I want to find out a closed form expression for the following series:

\[\sum_{k=1}^{\infty}k^{1-\alpha}\]

Of course this would be function of $\alpha$ . Actually in my case $\alpha>2$ . Any help is greatly appreciated.

#Calculus

Easy Math Editor

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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}`	$\boxed{123}$

Comments

Riemann Zeta function for $\alpha >2$ ! *except for that first zero term which as Sudeep Salgia said the value becomes infinity.

Kartik Sharma - 6 years, 5 months ago

Don't you think that for $\alpha >2$ and $k=0$ the value of the expression becomes infinity. ?

Sudeep Salgia - 6 years, 5 months ago

Thanks. That is a typo. Sum starts from $k=1$ .

Snehal Shekatkar - 6 years, 5 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}`	$\boxed{123}$