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24. $\lim _{n \rightarrow \infty}\left[\frac{1}{1-n^{2}}+\frac{2}{1-n^{2}}+\ldot...

Jan 30, 2024

24.

\lim _{n \rightarrow \infty}\left[\frac{1}{1-n^{2}}+\frac{2}{1-n^{2}}+\ldots \ldots \ldots+\frac{n}{1-n^{2}}\right]

is equal to (a) 0 (b) 2 (c)

1+e^{-1}

. (d) None of these.

Solution by Steps

step 2

The sum of the first n natural numbers is

\frac{n(n+1)}{2}.

step 3

Substitute the sum into the limit:

\lim_{n \to \infty} \frac{\frac{n(n+1)}{2}}{1 - n^2}.

step 4

Simplify the expression:

\lim_{n \to \infty} \frac{n(n+1)}{2(1 - n^2)}.

step 5

Divide numerator and denominator by

n^2

to get:

\lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2(\frac{1}{n^2} - 1)}.

step 6

n

approaches infinity,

\frac{1}{n}

and

\frac{1}{n^2}

approach 0, so the limit simplifies to:

\lim_{n \to \infty} \frac{1}{-2} = -\frac{1}{2}.

Key Concept

Limit of a sum involving a series as n approaches infinity

Explanation

The key concept is to recognize that the sum of the series can be simplified using the formula for the sum of the first n natural numbers and then taking the limit as n approaches infinity. By simplifying the expression and dividing by the highest power of n in the denominator, we can find the limit of the series.