Question

Pls answer this question
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Answer 1

Step-by-step explanation:

Recall that for any decreasing function f:R→R and any N>1 we have

∫1N+1f(x)dx≤∑k=1Nf(k)≤∫0Nf(x)dx

After substitutions N=n2, f(x)=n/(n2+x2) and simple computations we have

arctann2+1n−arctan1n≤∑k=1n2nn2+k2≤arctann

Lets take a limit n→∞, then from sandwich lemma it follows

limn→∞∑k=1n2nn2+k2=π2

P.S. First solution was not rigor enough.