Collection of Maths Problems

Ratio test

Task number: 2932

• Variant 1

  \(\displaystyle \sum_{n=1}^{\infty}\frac1{n!} \).

• Hint

  The expression \(n!\) becomes simpler if we study its ratio \(\frac{(n+1)!}{n!}=n+1\).

• Resolution

  We use the ratio test:

  \(\displaystyle \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{\frac1{(n+1)!}}{\frac1{n!}} =\lim_{n\to\infty}\frac1{n+1}=0<1. \)

  It would also be possible to use the comparison test, because whenever \(n\ge 2\), \(\frac1{n!}\le\frac1{n(n-1)}\).

• Result

  The series converges.

• Variant 2

  \(\displaystyle \sum\limits_{n=1}^{\infty} \frac{2^n}{n!}. \)

• Resolution

  \(\displaystyle \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}} =\lim_{n\to\infty}\frac{2}{n+1}=0<1. \)

• Result

  The series converges.

• Variant 3

  \(\displaystyle \sum\limits_{n=1}^{\infty} \frac{n!}{2^{n^2}}. \)

• Resolution

  \(\displaystyle \lim_{n\to\infty}\frac{a_{n+1}}{a_n}= \lim_{n\to\infty}\frac{\frac{(n+1)!}{2^{(n+1)^2}} }{\frac{n!}{2^{n^2}}}= \lim_{n\to\infty}\frac{n+1}{2^{2n+1}} \le \lim_{n\to\infty}\frac{2n}{2^{2n+1}} \le \lim_{n\to\infty}\frac{n}{2^{2n}} \le \lim_{n\to\infty}\frac{n}{2^n} =0<1 \).

• Result

  The series converges.

• Variant 4

  \(\displaystyle \sum\limits_{n=1}^{\infty} \binom{2n}n \frac{1}{5^n}. \)

• Resolution

  \(\displaystyle \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\binom{2n+2}{n+1} \frac{1}{5^{n+1}}}{\binom{2n}n \frac{1}{5^n}}= \lim_{n \to \infty}\frac{\frac{(2n+2)!}{(n+1)!(n+1)!}\cdot \frac{1}{5^{n+1}}}{\frac{2n!}{n!n!}\cdot\frac1{5^n}} = \lim_{n \to \infty} \frac{(2n + 2)(2n + 1)}{5(n+1)(n+1)} = \frac{4}{5}<1 \)

• Result

  The series converges.

• Variant 5

  \(\displaystyle \sum_{n=1}^{\infty}\frac{2^nn!}{n^n} \).

• Resolution

  \(\displaystyle \lim_{n\to\infty}\frac{a_{n+1}}{a_n}= \lim_{n\to\infty}\frac{\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}}{\frac{2^nn!}{n^n}}= \lim_{n\to\infty}\frac{2(n+1)n^n}{(n+1)^{n+1}}= 2 \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n=\\ =2 \lim\limits_{n\to\infty}\left(1-\frac1{n+1}\right)^{n+1}\left(1-\frac1{n+1}\right)^{-1}= 2e^{-1}\cdot1<1\).

• Result

  The series converges.

• Variant 6

  \(\displaystyle \sum_{n=1}^{\infty}\frac{3^nn!}{n^n} \).

• Resolution

  \(\displaystyle \lim_{n\to\infty}\frac{a_{n+1}}{a_n}= \lim_{n\to\infty}\frac{\frac{3^{n+1}(n+1)!}{(n+1)^{n+1}}}{\frac{3^nn!}{n^n}}= \lim_{n\to\infty}\frac{3(n+1)n^n}{(n+1)^{n+1}}= 3 \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n=\\ =3 \lim\limits_{n\to\infty}\left(1-\frac1{n+1}\right)^{n+1}\left(1-\frac1{n+1}\right)^{-1}= 3e^{-1}\cdot1>1\).

• Result

  The series diverges.

• Variant 7

  \(\displaystyle \sum_{n=1}^{\infty}\frac{(2n)!}{(n!)^2} \).

• Resolution

  \(\displaystyle \lim_{n\to\infty}\frac{a_{n+1}}{a_n}= \lim_{n\to\infty}\frac{\frac{(2n+2)!}{((n+1)!)^2}}{\frac{(2n)!}{(n!)^2}}= \lim_{n\to\infty}\frac{(2n+1)(2n+2)}{(n+1)^2}= \lim_{n\to\infty}\frac{(2+n^{-1})(2+2n^{-1})}{(1+n^{-1})^2}=4>1 \)

• Result

  The series diverges.

• Variant 8

  \(\displaystyle \sum_{n=1}^{\infty}\frac{n^n}{(2n)!}\).

• Resolution

  \(\displaystyle \lim_{n\to\infty}\frac{a_{n+1}}{a_n}= \lim_{n\to\infty}\frac{\frac{{n+1}^{n+1}}{(2n+2)!}}{\frac{n^n}{(2n)!}}= \lim_{n\to\infty}\frac{{n+1}^{n+1}}{{n^n}(2n+1)(2n+2)}= \lim_{n\to\infty}\frac{1}{4n+2}\left(1+\frac1n\right)^n=0 \cdot e<1 \)

• Result

  The series converges.