Question

Suppose you wanted to use the limit comparison test on $\sum_{n=1}^{\infty} \frac{1}{e^{n}-\ln n}$. Which of the following is the best series to compare $\sum_{n=1}^{\infty} \frac{1}{e^{n}-\ln n}$ to? (A) $\sum_{n=1}^{\infty} e^{n}$ (B) $\sum_{n=1}^{\infty} \frac{1}{n}$ (C) $\sum_{n=1}^{\infty} \frac{1}{\ln n}$ (D) $\sum_{n=1}^{\infty} \frac{-1}{\ln n}$ (E) None of above.

Name: suppose you wanted to use the limit comparison test on sumn1infty frac1en ln n which of the following is the best series to compare sumn1infty frac1en ln n to a sumn1infty en b sumn1inft 57869
Uploaded: 2022-01-04T05:46:38-08:00
Duration: 6 min 19 s
Channel: Uma Kumari
Description: suppose you wanted to use the limit comparison test on sumn1infty frac1en ln n which of the following is the best series to compare sumn1infty frac1en ln n to a sumn1infty en b sumn1inft 57869

          Suppose you wanted to use the limit
comparison test on $\sum_{n=1}^{\infty} \frac{1}{e^{n}-\ln n}$.
Which of the following is the best series
to compare $\sum_{n=1}^{\infty} \frac{1}{e^{n}-\ln n}$ to?
(A) $\sum_{n=1}^{\infty} e^{n}$
(B) $\sum_{n=1}^{\infty} \frac{1}{n}$
(C) $\sum_{n=1}^{\infty} \frac{1}{\ln n}$
(D) $\sum_{n=1}^{\infty} \frac{-1}{\ln n}$
(E) None
of
above.

$suppose you wanted to use the limit comparison test on sumn1infty frac1en ln n which of the following is the best series to compare sumn1infty frac1en ln n to a sumn1infty en b sumn1inft 57869$

Added by Sandra N.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Uma Kumari

01/04/2022

Step 1

The Limit Comparison Test states that if we have two series $\sum a_n$ and $\sum b_n$ with positive terms, and if $\lim_{n \to \infty} \frac{a_n}{b_n} = L$ where $L$ is a finite, positive number ($0 < L < \infty$), then either both series converge or both series Show more…

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Suppose you wanted to use the limit comparison test on $\sum_{n=1}^{\infty} \frac{1}{e^{n}-\ln n}$. Which of the following is the best series to compare $\sum_{n=1}^{\infty} \frac{1}{e^{n}-\ln n}$ to? (A) $\sum_{n=1}^{\infty} e^{n}$ (B) $\sum_{n=1}^{\infty} \frac{1}{n}$ (C) $\sum_{n=1}^{\infty} \frac{1}{\ln n}$ (D) $\sum_{n=1}^{\infty} \frac{-1}{\ln n}$ (E) None of above.