Investigating Convergence of a Series

[sr1923]

Hi, could someone help me identify an appropriate test to determine whether the following series converges or not. Thank you

 

[blamocur]

Show that [imath]a_n > \frac{1}{n}[/imath]

 

[mario99]

Hi, could someone help me identify an appropriate test to determine whether the following series converges or not. Thank youView attachment 38421

[imath]\displaystyle \sum_{n=1}^{\infty}\frac{(2n)!}{2^{2n}(n!)^{2}} \ \ \approx \ \ \frac{1}{\sqrt{\pi}}\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}[/imath]


What about the integral test?

You can also use the ratio test directly on the original sum. But you will have to replace the factorial by using the sterling formula while calculating the limit.

 

[lookagain]

But you will have to replace the factorial by using the sterling formula while calculating the limit.

Are you referring to these two factorial forms in the ratio test?

Spoiler

\(\displaystyle (2n)! \ \ and \ \ (n!)^2\)

------‐------------------------

For \(\displaystyle \ a_{n + 1}, \ \) the respective parts are

\(\displaystyle (2(n + 1))! \ = \ (2n + 2)! \ = \ (2n + 2)(2n + 1)(2n)! \ \ \) and

\(\displaystyle ((n + 1)!)^2 \ = \ ((n + 1)n!)^2 \ = \ (n + 1)^2(n!)^2 \)

The two expressions at the end of the chains of equals can be
substituted into the larger expression for the ratio test, and the
\(\displaystyle (2n)! \ \ and \ \ (n!)^2 \ \) factors will divide out between the numerator
and the denominator. (There are also the other expressions
involving the powers of two.)

However, if the limit of (the absolute value) of \(\displaystyle \ \dfrac{\ a_{n + 1}}{a_n} \ \) equals 1,
then this test is inconclusive.

 

[mario99]

Are you referring to these two factorial forms in the ratio test?

Spoiler

\(\displaystyle (2n)! \ \ and \ \ (n!)^2\)

------‐------------------------

For \(\displaystyle \ a_{n + 1}, \ \) the respective parts are

\(\displaystyle (2(n + 1))! \ = \ (2n + 2)! \ = \ (2n + 2)(2n + 1)(2n)! \ \ \) and

\(\displaystyle ((n + 1)!)^2 \ = \ ((n + 1)n!)^2 \ = \ (n + 1)^2(n!)^2 \)

The two expressions at the end of the chains of equals can be
substituted into the larger expression for the ratio test, and the
\(\displaystyle (2n)! \ \ and \ \ (n!)^2 \ \) factors will divide out between the numerator
and the denominator. (There are also the other expressions
involving the powers of two.)

However, if the limit of (the absolute value) of \(\displaystyle \ \dfrac{\ a_{n + 1}}{a_n} \ \) equals 1,
then this test is inconclusive.

Yeah. He first may use the ratio test as usual. When he gets 1, he should try something else which is the integral test on the new sum.