Question

$\lim_{n\to\infty} [\frac{1}{n} + \frac{n}{(n+1)^2} + \frac{n}{(n+2)^2} + \dots + \frac{n}{(2n-1)^2}]$ is equal to :

• A. 1
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Hint:

To convert a limit of a sum into a definite integral, manipulate the expression into the form $\lim_{n\to\infty} \frac{1}{n} \sum f(\frac{r}{n})$. In this form, replace $\frac{1}{n}$ with $dx$, $\frac{r}{n}$ with $x$, and the summation with an integral sign. The limits of integration are typically from 0 to 1 if the sum is from $r=0$ to $n-1$.

Answer 1

The Correct Option is D

Given: \[ \lim_{n\to\infty}\left[ \frac{1}{n}+\frac{n}{(n+1)^2}+\cdots+\frac{n}{(2n-1)^2} \right] \] Step 1: Write as summation \[ = \lim_{n\to\infty} \sum_{r=0}^{n-1} \frac{n}{(n+r)^2} \] Step 2: Convert to Riemann sum \[ = \lim_{n\to\infty}\frac{1}{n}\sum_{r=0}^{n-1} \frac{1}{(1+\frac{r}{n})^2} \] Step 3: Convert to integral \[ = \int_0^1 \frac{1}{(1+x)^2}\,dx \] Step 4: Integrate \[ = \left[-\frac{1}{1+x}\right]_0^1 = 1-\frac{1}{2} = \frac{1}{2} \] \[ \boxed{\frac{1}{2}} \]