Question

7. (10 points) Use the Integral Test, Comparison Test, or Limit Comparison Test to determine whether the series converge or diverge. (a) $sum_{n=1}^{infty} frac{n+4^n}{n+5^n}$ (b) $sum_{n=1}^{infty} frac{ln n}{n^3}$ (c) $sum_{n=1}^{infty} frac{n^3+2n}{n^4+3n^3+2n}$ (d) $sum_{n=1}^{infty} frac{sqrt{4n^3+1}}{6n^3+5n^2+3}$

          7. (10 points) Use the Integral Test, Comparison Test, or Limit Comparison Test to determine whether the series converge or diverge.
(a) $sum_{n=1}^{infty} frac{n+4^n}{n+5^n}$
(b) $sum_{n=1}^{infty} frac{ln n}{n^3}$
(c) $sum_{n=1}^{infty} frac{n^3+2n}{n^4+3n^3+2n}$
(d) $sum_{n=1}^{infty} frac{sqrt{4n^3+1}}{6n^3+5n^2+3}$

$7. (10 points) Use the Integral Test, Comparison Test, or Limit Comparison Test to determine whether the series converge or diverge. (a) sumn=1^infty fracn+4^nn+5^n (b) sumn=1^infty fracln nn^3 (c) sumn=1^infty fracn^3+2nn^4+3n^3+2n (d) sumn=1^infty fracsqrt4n^3+16n^3+5n^2+3$

Added by Cassandra R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Dominique Jan Tan

05/16/2023

Step 1

By comparing the terms, we can see that $\frac{n+4}{n^2 + 2n + 5} < \frac{n+4}{n^2}$ for all $n$. Since $\sum_{n=1}^{\infty} \frac{n+4}{n^2}$ converges (as it is a p-series with $p=2$), the given series also converges by the Comparison Test. ** Show more…

Show all steps

Thanks for your feedback!

7. Use the Integral Test, Comparison Test, or Limit Comparison Test to determine whether the series converge or diverge. (a) sum_{n=1}^{infinity} (n+4^n)/(n+5^n) (b) sum_{n=1}^{infinity} (ln n)/(n^3) (c) sum_{n=1}^{infinity} (n^3+2n)/(n^4+3n^3+2n) (d) sum_{n=1}^{infinity} (sqrt(4n^3+1))/(6n^3+5n^2+3)