Find the value of limn-1-2-3-n-n2

The correct option is B We have, lim_n→∞1+2+3+....n/n^2 Here, numerator is sum of first n natural number we know, ∑ n=n(n+1)/2 So, lim_n→∞n(n+1)/2/n^2 ...

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Standard XI

Mathematics

Standard Limits to Remove Indeterminate Form

Find the valu...

Question

Find the value of $lim n \to \infty \frac{1 + 2 + 3 + . . . . n}{n^{2}}$

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Solution

The correct option is D

We have, $lim n \to \infty \frac{1 + 2 + 3 + . . . . n}{n^{2}}$

Here, numerator is sum of first n natural number

we know, $\sum n = \frac{n (n + 1)}{2}$

So, $lim n \to \infty \frac{\frac{n (n + 1)}{2}}{n^{2}}$

= $lim n \to \infty \frac{(n + 1)}{2 n}$

= $lim n \to \infty (\frac{1}{2} + \frac{1}{2 n})$

$= \frac{1}{2} + \frac{1}{\infty}$

$= \frac{1}{2} + 0$

$= \frac{1}{2}$

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Find the value of limn→∞1+2+3+....nn2

The correct option is D We have, limn→∞1+2+3+....nn2 Here, numerator is sum of first n natural number we know, ∑n=n(n+1)2 So, limn→∞n(n+1)2n2 = limn→∞(n+1)2n = limn→∞(12+12n) =12+1∞ =12+0 =12

Find the value of $lim n \to \infty \frac{1 + 2 + 3 + . . . . n}{n^{2}}$