Lim- k-1n k-n2-k2 is equals to -Roorkee 1999-

Let I= lim_π→∞∑ _k=1^n k/n^2+k^2=lim_π→∞∑ _k=1^n 1/n ( k/n )/1+ ( k/n )^2 I = ∫_0^1 x/1+x^2dx=1/2 [log (1+x^2)]^1_0 =1/2[log 2] ...

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Definite Integral as Limit of Sum

limπ→∞∑ k=1n ...

Question

$l i m_{π \to \infty} \sum_{k = 1}^{n} \frac{k}{n^{2} + k^{2}}$ is equals to [Roorkee 1999]

\frac{1}{2} l o g 2

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l o g 2

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\frac{π}{4}

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\frac{π}{2}

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Solution

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$l i m_{π \to \infty} \sum_{k = 1}^{n} \frac{k}{n^{2} + k^{2}}$ is equals to [Roorkee 1999]

S_{n} = \sum_{k = 1}^{n} \frac{n}{n^{2} + k n + k^{2}}, T_{n} = n - 1 \sum k = 0 \frac{n}{n^{2} + k n + k^{2}}

n = 1, 2, 3...

S_{n} = n \sum k = 1 \frac{n}{n^{2} + k n + k^{2}}

T_{n} = n - 1 \sum k = 0 \frac{n}{n^{2} + k n + k^{2}}

n = 1, 2, 3, \dots

S_{n} = n \sum k = 1 \frac{n}{n^{2} + k n + k^{2}}

T_{n} = n - 1 \sum k = 0 \frac{n}{n^{2} + k n + k^{2}}

n = 1, 2, 3, . .

S_{n} = \sum_{k = 1}^{n} \frac{n}{n^{2} + k n + k^{2}} a n d T_{n} = \sum_{k = 0}^{n - 1} \frac{n}{n^{2} + k n + k^{2}}

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