Question
$\lim _{n \rightarrow \infty}\left(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{n(n+1)}\right)$ is equal to(A) 1(B) $-1$(C) 0(D) None of these
Step 1
Step 1: We are given the series $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{n(n+1)}$ and we need to find its limit as $n$ approaches infinity. Show more…
Show all steps
Your feedback will help us improve your experience
Abhijith V and 88 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
$\lim _{n \rightarrow \infty} \frac{1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+n(n+1)}{n^{3}}$ is equal to (A) 1 (B) $-1$ (C) $\frac{1}{3}$ (D) None of these
$\lim _{n \rightarrow \infty}\left[\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\ldots+\frac{1}{(2 n+1)(2 n+3)}\right]$ is equal to (A) 1 (B) $\frac{1}{2}$ (C) $-\frac{1}{2}$ (D) None of these
$\lim _{n \rightarrow \infty}\left\{\frac{1}{1-n^{2}}+\frac{2}{1-n^{2}}+\ldots .+\frac{n}{1-n^{2}}\right\}$ is equal to (a) 0 (b) $-\frac{1}{2}$ (c) $\frac{1}{2}$ (d) none of these
Watch the video solution with this free unlock.
EMAIL
PASSWORD