Determine whether the series converges, and if so, find its sum.
(1) \(\sum_{n=1}^{\infty} 3^{-n} 8^{n+1}\)
\((2) \sum_{n=2}^{\infty} \frac{1}{n(n-1)}\)
(3) \(\sum_{n=0}^{\infty}(-3)\left(\frac{2}{3}\right)^{2 n}\)
(4) \(\sum_{n=1}^{\infty} \frac{1}{e^{2 n}}\)
(5) \(\sum_{n=1}^{\infty} \ln \frac{n}{n+1}\)
(6) \(\sum_{n=1}^{\infty}[\arctan (n+1)-\arctan n]\)
(7) \(\sum_{n=1}^{\infty} \ln \left(\frac{n^{2}+4}{2 n^{2}+1}\right)\)
(8) \(\sum_{n=1}^{\infty} \frac{1+2^{n}}{3^{n}}\)
(9) \(\sum_{n=1}^{\infty}\left[\cos \frac{1}{n^{2}}-\cos \frac{1}{(n+1)^{2}}\right]\)
Determine whether the series converges or diverges.(1) \(\sum_{n=1}^{\infty} \frac{e^{1 / n}}{n^{2}}\)(2) \(\sum_{n=1}^{\infty}\left(\frac{2}{\sqrt{n}}+\frac{(-1)^{n}}{3^{n+1}}\right)\)(3) \(\sum_{n=1}^{\infty} \frac{5-2 \sin n}{n}\)(4) \(\sum_{n=1}^{\infty} \frac{3+\cos n}{n^{3 / 2}}\)(5) \(\sum_{n=0}^{\infty} \frac{\sqrt{n^{2}+2}}{n^{4}+n^{2}+5}\)(6) \(\sum_{n=1}^{\infty=1}\left(1+\frac{1}{n}\right)^{n}\)(7) \(\sum_{n=1}^{\infty} \frac{n+1}{n 2^{n}}\)(8) \(\sum_{n=1}^{\infty} \frac{\arctan n}{n^{4}}\)(9) \(\sum_{n=1}^{\infty} n \sin \frac{1}{n}\)
1. Determine whether the series converges or diverges.$$ \sum_{k=1}^{\infty} \frac{\ln (k)}{k} $$convergesdiverges2.Test the series for convergence or divergence.$$ \sum_{n=1}^{\infty}(-1)^{n} \sin \left(\frac{3 \pi}{n}\right) $$convergesdiverges
Determine whether the given series converges or diverges. Fully justify your answe(a) \(\sum_{n=2}^{\infty} \frac{1}{\sqrt{n} \ln n}\)(b) \(\sum_{n=1}^{\infty} \cos \left(\frac{1}{n^{2}}\right)\)(c) \(\sum_{n=1}^{x} \frac{(2 n) !}{5^{n} n ! n t}\)
Determine whether the series is convergent or divergent.$$ \sum_{n=1}^{\infty}\left(\frac{8}{e^{n}}+\frac{4}{n(n+1)}\right) $$convergentdivergentIf it is convergent, find its sum. (If the quantity diverges, enter DIVERGES.)
7. Use the Alternating Series Test to determine the convergence or divergence of the series a) \(\sum_{n=1}^{\infty} \frac{(-1)^{n} \sqrt{n}}{2 n+1}\)b) \(\sum_{n=1}^{\infty} \frac{(-1)^{n} n}{2 n-1}\)8. Use the Ratio Test or the Root Test to determine the convergence or divergence of the seriesa) \(\sum_{n=0}^{\infty}\left(\frac{4 n-1}{5 n+7}\right)^{n}\)b) \(\sum_{n=0}^{\infty} \frac{\pi^{n}}{n !}\)
Write out the first five terms of the sequence with, \(\left[\frac{\ln(n)}{n+1}\right]_{n=1}\), determine whether the sequence converges, and if so find its limit. Enter the following information for \(a_{n}=\frac{\ln (n)}{n+1}\). \(a_{1}=\) \(a_{2}=\) \(a_{3}=\) \(a_{4}=\) \(a_{5}=\) \(\lim_{n \rightarrow \infty} \frac{\ln (n)}{n+1}=\) (Enter DNE if limit Does Not Exist.) Does the sequence converge (Enter "yes" or "no").
If the series \(\sum_{n=1}^{\infty} a_{n}\) converges and \(a_{n}>0\) for all \(n\), which of the following must be true?(A) \(\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=0\)(B) \(\left|a_{n}\right|<1\)for all \(n\)(C) \(\sum_{n=1}^{\infty} a_{n}=0\)(D) \(\sum_{n=1}^{\infty} n a_{n}\) diverges.(E) \(\sum_{n=1}^{\infty} \frac{a_{n}}{n}\) converges.
Determine whether the sequence converges or diverges. If it converges, find the limit. (If an answer does not exist, enter DNE.) $$ \begin{gathered} a_{n}=\ln \left(2 n^{2}+6\right)-\ln \left(n^{2}+6\right) \\ \lim _{n \rightarrow \infty} a_{n}= \end{gathered} $$
Question 21 Indicate whether the series, \sum_{n=1}^{\infty} \frac{5}{2n^2 + 4n+ 3} converges or diverges. Select one: a. Converges b. Diverges
find an expression for the area of the region under the graph f(x)=x^4 on the interval [1,7]. use right-Hand endpoints as sample points choices1. area \(=\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(1+\frac{7 i}{n}\right)^{4} \frac{7}{n}\)2. area \(=\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(1+\frac{9 i}{n}\right)^{4} \frac{6}{n}\)3. area \(=\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(1+\frac{6 i}{n}\right)^{4} \frac{6}{n}\)4. area \(=\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(1+\frac{7 i}{n}\right)^{4} \frac{6}{n}\)5. area \(=\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(1+\frac{6 i}{n}\right)^{4} \frac{7}{n}\)6. area \(=\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(1+\frac{9 i}{n}\right)^{4} \frac{7}{n}\)