Collection of Maths Problems

Various series

Task number: 2934

• Variant 1

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

• Resolution

  By the comparison test:

  \(\displaystyle \sum_{n=1}^{\infty}\frac{n+1}{n(n+2)}\ge \sum_{n=1}^{\infty}\frac{n}{3n^2}\ge \frac13\sum_{n=1}^{\infty}\frac{1}{n} \).

• Result

  The series diverges.

• Variant 2

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

• Resolution

  \(\displaystyle \lim_{n\to\infty}\sqrt\frac{n-1}{2n}=\frac{\sqrt{2}}2>0\).

• Result

  The series diverges.

• Variant 3

  \(\displaystyle \sum_{n=1}^{\infty}\frac{n+1}{(n+1)\sqrt{n+1}-1} \)

• Resolution

  By the comparison test:

  \(\displaystyle \sum_{n=1}^{\infty}\frac{n+1}{(n+1)\sqrt{n+1}-1} = \sum_{n=2}^{\infty}\frac{n}{n\sqrt{n}-1} \ge \sum_{n=2}^{\infty}\frac1{\sqrt{n}} \)

• Result

  The series diverges.

• Variant 4

  \(\displaystyle \sum_{n=1}^{\infty}\left(\frac{1+n^2}{1+n^3}\right)^2 \).

• Resolution

  By the comparison test:

  \(\displaystyle \sum_{n=1}^{\infty}\left(\frac{1+n^2}{1+n^3}\right)^2 \le \sum_{n=1}^{\infty}\frac{n^4+2n^2+1}{n^6} \le 4 \sum_{n=1}^{\infty}\frac{1}{n^2} \).

• Result

  The series converges.

• Variant 5

  \(\displaystyle \sum_{n=1}^{\infty}\frac1{(n+3)\sqrt{n}} \).

• Resolution

  By the comparison test:

  \(\displaystyle \sum_{n=1}^{\infty}\frac1{(n+3)\sqrt{n}}\le \sum_{n=1}^{\infty} n^{-\frac32} \).

• Result

  The series converges.

• Variant 6

  \(\displaystyle \sum_{n=1}^{\infty}\frac{\sqrt n}{n^3+1} \).

• Resolution

  By the comparison test:

  \(\displaystyle \sum_{n=1}^{\infty}\frac{\sqrt n}{n^3+1}\le \sum_{n=1}^{\infty} n^{-2} \).

• Result

  The series converges.

• Variant 7

  \(\displaystyle \sum_{n=1}^{\infty}\sqrt[n]{\frac1{1000}} \).

• Resolution

  \(\displaystyle \lim_{n\to\infty}\sqrt[n]{\frac1{1000}}=1 \).

• Result

  The series diverges.

• Variant 8

  \(\displaystyle \sum_{n=1}^{\infty}\frac{2n-1}{\sqrt{2^n}} \)

• Resolution

  By the root test:

  \(\displaystyle \lim_{n\to\infty}\sqrt[n]{\frac{2n-1}{\sqrt{2^n}}}= \frac{\lim\limits_{n\to\infty}\sqrt[n]{2n-1}}{\sqrt{2}}= \frac{\sqrt{2}}2<1 \).

  Notice that

  \(\displaystyle 1\le \lim_{n\to\infty}\sqrt[n]{2n-1}\le \lim_{n\to\infty}\sqrt[n]{3n}= \lim_{n\to\infty}\sqrt[n]{3}\cdot\lim_{n\to\infty}\sqrt[n]{n}=1{\cdot} 1=1 \).

• Result

  The series converges.

• Variant 9

  \(\displaystyle \sum_{n=1}^{\infty}\frac1{n\cdot3^n} \).

• Resolution

  By the comparison test \(\displaystyle \sum_{n=1}^{\infty}\frac1{n\cdot3^n} \le \sum_{n=1}^{\infty}\frac1{n^2} \).

  Or by the root test: \(\displaystyle \lim_{n\to\infty}\sqrt[n]{\frac1{n\cdot3^n}}=\frac1{3\lim\limits_{n\to\infty}\sqrt[n]{n}}=\frac13<1 \).

• Result

  The series converges.

• Variant 10

  \(\displaystyle \sum_{n=1}^{\infty}\frac{n^5}{2^n+3^n} \).

• Resolution

  First the comparison test, then the root test:

  \(\displaystyle \sum_{n=1}^{\infty}\frac{n^5}{2^n+3^n}\le \sum_{n=1}^{\infty}\frac{n^5}{2^n} \)

  \(\displaystyle \lim_{n\to\infty}\sqrt[n]{\frac{n^5}{2^n}}= \frac{\left(\lim\limits_{n\to\infty}\sqrt[n]{n}\right)^5}{2}=\frac12 \)

  The second series converges, so the first does also.

• Result

  The series converges.

• Variant 11

  \(\displaystyle \sum_{n=1}^{\infty}\frac{1!+2!+3!+…+n!}{(2n)!} \).

• Resolution

  By the comparison test:

  \(\displaystyle \sum_{n=1}^{\infty}\frac{1!+2!+3!+…+n!}{(2n)!} \le \sum_{n=1}^{\infty}\frac{n\cdot n!}{(2n)!} \le \sum_{n=1}^{\infty}\frac{n}{(n+1)(n+2)(n+3)} \le \sum_{n=1}^{\infty}\frac{1}{n^2} \).

• Result

  The series converges.

• Variant 12

  \(\displaystyle \sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}\right)^n \).

• Resolution

  \(\displaystyle \lim_{n\to\infty}{\left(\frac{n+1}{n+2}\right)^n}=e^{-1}>0 \).

• Result

  The series diverges.

• Variant 13

  \(\displaystyle \sum_{n=1}^{\infty}\left(\frac{1+\cos n}{2+\cos n}\right)^n \).

• Resolution

  The cosine function has values in the range \([-1{,}1]\).

  For all \(t\in[-1{,}1]\), \(\displaystyle \frac{1+t}{2+t}=1-\frac{1}{2+t}\leq 1-\frac{1}{2+1} = \frac{2}{3} \).

  We proceed using the comparison test:

  \(\displaystyle \sum_{n=1}^{\infty}\left(\frac{1+\cos n}{2+\cos n}\right)^n\leq \sum_{n=1}^{\infty}\left(\frac{2}{3} \right)^n \).

• Result

  The series converges.

• Variant 14

  \(\displaystyle \sum_{n=1}^{\infty}\frac{2+\cos n}{n+\ln n} \)

• Resolution

  By the comparison test:

  \(\displaystyle \sum_{n=1}^{\infty}\frac{2+\cos n}{n+\ln n} \ge \sum_{n=1}^{\infty}\frac{1}{2n} \)

• Result

  The series diverges.

• Variant 15

  \(\displaystyle \sum_{n=1}^{\infty}\frac{\sin n}{n(n+2)} \)

• Resolution

  By the comparison test:

  \(\displaystyle \sum_{n=1}^{\infty}\frac{\sin n}{n(n+2)} \le \sum_{n=1}^{\infty}\frac{1}{n^2} \)

• Result

  The series converges.

• Variant 16

  \(\displaystyle \sum_{n=1}^{\infty}\sin\frac1n \).

• Resolution

  By the comparison test:

  \(\displaystyle \sum_{n=1}^{\infty}\sin\frac1n \ge \sum_{n=1}^{\infty}\frac1{2n} \ge \frac12\sum_{n=1}^{\infty}\frac1{n} \).

• Result

  The series diverges.

• Variant 17

  \(\displaystyle \sum_{n=1}^{\infty}\sin\frac1{n^2} \).

• Resolution

  By the comparison test:

  \(\displaystyle \sum_{n=1}^{\infty}\sin\frac1{n^2} \le \sum_{n=1}^{\infty}\frac1{n^2} \).

• Result

  The series converges.

• Variant 18

  \(\displaystyle \sum_{n=1}^{\infty}e^{\sqrt[-3]n} \).

• Resolution

  We use the fact that \(\displaystyle \lim_{k\to\infty}\frac{k^6}{e^k}=0 \), so there exists \(k_0\in\mathbb N\) such that for all natural numbers \(k\geq k_0\), \(\displaystyle \frac{k^6}{e^k}\leq 1. \)

  For \(n\geq n_0:=k_0^3\) using an elementary estimate of the integer part \(x-1\leq \lfloor x\rfloor\leq x\) we know that for \(k=\lfloor\root{3}\of{n}\rfloor\),

  \(\displaystyle \frac{n^2}{e^{\root{3}\of{n}}}= \frac{(\root{3}\of{n})^6}{e^{\root{3}\of{n}}}\leq \frac{(\lfloor\root{3}\of{n}\rfloor+1)^6}{e^{\lfloor\root{3}\of{n}\rfloor}}\leq \frac{(2k)^6}{e^k}\leq 2^6. \)

  So for \(n\geq n_0\) we have the estimate \(\displaystyle e^{-\root{3}\of{n}}=\frac{1}{e^{\root{3}\of{n}} }\leq \frac{2^6}{n^2} . \)

  The series \(\displaystyle\sum_{n=1}^{\infty}\frac{2^6}{n^2}\) converges, and so by the comparison test for series we know that the original series converges as well.

• Result

  The series converges.

• Variant 19

  \(\displaystyle \sum_{n=1}^{\infty}\frac{\ln n}{n} \)

• Resolution

  By the comparison test:

  \(\displaystyle \sum_{n=1}^{\infty}\frac{\ln n}{n}\ge \sum_{n=1}^{\infty}\frac{1}{n} \)

• Result

  The series diverges.

• Variant 20

  \(\displaystyle \sum_{n=2}^{\infty}\frac{\sqrt{n^2+1}-n}{\log^2 n}\ . \)

• Resolution

  First we transform the numerator to the difference of roots \(\sqrt{n^2+1}-\sqrt{n^2}\), and so we can expand the entire expression

  \(\displaystyle a_n=\frac{\sqrt{n^2+1}-n}{\log^2 n}=\frac{\sqrt{n^2+1}-\sqrt{n^2}}{\log^2 n}\frac{\sqrt{n^2+1}+\sqrt{n^2}}{\sqrt{n^2+1}+\sqrt{n^2}} =\frac{(n^2+1)-n^2}{(\sqrt{n^2+1}+\sqrt{n^2})\log^2 n}. \)

  Now we can easily guess that this sequence behaves roughly like the sequence \(\displaystyle b_n=\frac{1}{n \log^2 n}. \)

  \(\displaystyle \lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\frac{1}{(\sqrt{n^2+1}+\sqrt{n^2})\log^2 n}}{\frac{1}{n \log^2 n}} =\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{1}{n^2}}+1}=\frac{1}{2}\in(0,\infty). \)

  The sequence \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n \log^2 n}\) converges, and so the given series also converges by the limit comparison test.

• Result

  The series converges.