Question

Please answer both questions

F(-1)*5 54+1 4. (17pts) The series 324 converges. Write an argument to determine what value the series k=1 converges to. 5. (17pts) Write an argument to determine whether the series & 2* '(k+1)! converges or diverges. kol

Answer 1

__Part 1)__

Here the given series is:

(-1)* 54+1 32k k=1

Since it is convergent so we need to find at what value it is converging.

Absolute convergence:

If $\sum \left | a_n \right |$ is converegent then $\sum a_n$ is absolutely convergent.

Hence:

$a_n=\sum_{k=1}^{\infty}\dfrac{(-1)^k\ 5^{k+1}}{3^{2k}}\\ And\\ a_{n+1}=\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}\ 5^{(k+1)+1}}{3^{2(k+1)}}=\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}\ 5^{k+2}}{3^{2k+2}}\\$

Therefore:

$\lim_{k\rightarrow \infty}\left | \dfrac{a_{n+1}}{a_n} \right |\\ =\lim_{k\rightarrow \infty}\left |\dfrac{\frac{(-1)^{k+1}\ 5^{k+2}}{3^{2k+2}}}{\frac{(-1)^k\ 5^{k+1}}{3^{2k}}} \right |\\ =\lim_{k\rightarrow \infty}\left |\dfrac{(-1)^k(-1)\ 5^k\cdot 5^2}{3^{2k}\ 3^2}\cdot \dfrac{3^{2k}}{(-1)^k\ 5^k\cdot 5} \right |\\ =\lim_{k\rightarrow \infty}\left |\dfrac{-5}{9} \right |\\ =\dfrac{5}{9}$

As it gives finite value so it is absolutely converging.

__Part 2)__

Here we have to find the given series is converging or diverging.

Given:

$\sum_{k=1}^{\infty}\dfrac{2^k}{(k+1)!}$

Now:

$a_n=\sum_{k=1}^{\infty}\dfrac{2^k}{(k+1)!}\\ a_{n+1}=\sum_{k=1}^{\infty}\dfrac{2^{k+1}}{((k+1)+1)!}=\sum_{k=1}^{\infty}\dfrac{2^k}{(k+2)!}\\$

Hence:

$\lim_{k\rightarrow \infty}\left | \dfrac{a_{n+1}}{a_n} \right |$

Therefore:

$\lim_{k\rightarrow \infty}\left | \dfrac{\frac{2^{k+1}}{(k+2)!}}{\frac{2^k}{(k+1)!}} \right |\\ =\lim_{k\rightarrow \infty}\left | \dfrac{2^k\cdot 2}{(k+2)!}\cdot \dfrac{(k+1)!}{2^k}\right |\\ =\lim_{k\rightarrow \infty}\left | \dfrac{2 (k+1)!}{(k+2)(k+1)!} \right |\\ =\lim_{k\rightarrow \infty}\left | \dfrac{2}{k\left ( 1+\dfrac{2}{k} \right )} \right |\\$

Hence by applying series ratio test we can see the given series is also converging.