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Limit of a sum in Wolfram|Alpha

Posted 10 years ago

Could anyone help me correcting my formula? I tried to enter the following limit of a sum in Wolfram alpha, but it doesn't seem to understand it correctly.

limit (sum (i/(i^2 + n^2)), i=1..n) as n->infinity

Any suggestions are very welcome!

POSTED BY: Wim Lamotte
4 Replies

The original poster was asking about beating this information out of Wolfram|Alpha.

I got part way by splitting the problem. First I did the sum,

 sum (i/(i^2 + n^2)), i=1 to n 

I then used the Copyable PlainText link to get the output into another W|A window. I wrote the Limit out fully.

Limit as n goes to infinity of 1/2 (-polygamma(0, 1-i n)-polygamma(0, i n+1)+polygamma(0, (1-i) n+1)+polygamma(0, (1+i) n+1))

That tried but ran out of time.

simplify (-polygamma(0, 1-i n)-polygamma(0, i n+1)+polygamma(0, (1-i)  n+1)+polygamma(0, (1+i) n+1))

Gave a sum of Harmonic numbers, which did not look nicer to deal with.

POSTED BY: Bruce Miller
Posted 10 years ago

Appreciate the joke. I should have applied FullSimplify before posting (or done a bit more thinking).

POSTED BY: Jim Baldwin

Obviously this is the general result for complex values of 1.

In the special case where 1 takes real values the above expression simplifies to Log(2)/2.

(Just could not resist the joke, sorry!)


POSTED BY: Henrik Schachner
Posted 10 years ago

Using Mathematica 10.0.1 the limit is

Limit of sum

which is approximately 0.346574.

POSTED BY: Jim Baldwin
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