Question
If $S_{n}=\sum_{k=1}^{n} \frac{k^{2}}{1+n^{3}}$ then $\lim _{n \rightarrow \infty} S_{n}$ equals(a) 2(b) $-\frac{1}{3}$(c) $\frac{1}{3}$(d) $-2$
Step 1
We can rewrite this sum as $S_{n}=\frac{1}{1+n^{3}}\sum_{k=1}^{n} k^{2}$. Show more…
Show all steps
Your feedback will help us improve your experience
Suman Saurav Thakur and 72 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
If $t_{r}=\frac{1^{2}+2^{2}+3^{2}+\ldots+r^{2}}{1^{3}+2^{3}+3^{3}+\ldots+r^{3}}$ and $S_{n}=\sum_{r=1}^{n}(-1)^{r} \cdot t_{r}$, then $\lim _{n \rightarrow \infty} \mathrm{S}_{n}$ is given by (A) $\frac{2}{3}$ (B) $-\frac{2}{3}$ (C) $\frac{1}{3}$ (D) $-\frac{1}{3}$
If $S_{m}=\sum_{r=1}^{n} t_{r}=\frac{1}{6} n(n+1)(n+2) \forall n \geq 1$, then $\lim _{n \rightarrow-\infty} \sum_{n=1}^{n} \frac{1}{t_{r}}$ is (a) 1 (b) $3 / 2$ (c) 2 (d) $5 / 2$
$\lim _{n \rightarrow \infty}\left(\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\ldots+\frac{1}{(2 n-1)(2 n+1)}\right)$ is (a) 0 (b) 1 (c) $\frac{1}{2}$ (d) 2
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD