No links please Q Evaluate : limn→∞n+1n2+12+n+2n2+22+n+3n2+32+ +1n - Maths - Integrals

Question

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Q Evaluate : lim n → ∞ n + 1 n 2 + 1 2 + n + 2 n 2 + 2
2 + n + 3 n 2 + 3 2 + + 1 n

Neha Sethi · Accepted Answer

Dear student
limn→∞n+1n2+12+n+2n2+22+n+3n2+32+
+1n=limn→∞∑n+rn2+r2=limn→∞∑1n1+rn1n21+r2n2=limn→∞∑n1+rn1+r2n2Therefore, replacing rn by x and n by dx, we getf(x)=1+x1+x2 and ∫f(x)dx=∫1+x1+x2dxLower Limit=limn→∞rn=limn→∞1n=0Upper limit=limn→∞rn=limn→∞nn=1Hence the required limit is∫011+x1+x2dx=tan-1x+12log1+x201=tan-11+12log1+12-tan-10+12log1+02=tan-1tanπ4+12log2-tan-1tan0+12log1=π4+12log2      log1=0
Regards

No links please Q. Evaluate : lim n → ∞ n + 1 n 2 + 1 2 + n + 2 n 2 + 2 2 + n + 3 n 2 + 3 2 + . . . . . . . . + 1 n

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Q. Evaluate : $\lim_{n \to \infty} [\frac{n + 1}{n^{2} + 1^{2}} + \frac{n + 2}{n^{2} + 2^{2}} + \frac{n + 3}{n^{2} + 3^{2}} + . . . . . . . . + \frac{1}{n}]$