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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!
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Posted 10 years ago
 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 10 years ago
 Appreciate the joke. I should have applied FullSimplify before posting (or done a bit more thinking).
Posted 10 years ago
 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!)Henrik
Posted 10 years ago
 Using Mathematica 10.0.1 the limit is which is approximately 0.346574.