Collection of Maths Problems

Comparison test

Task number: 2930

• Variant 1

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

• Resolution

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

  We use the fact that the harmonic series diverges.

• Result

  The series diverges.

• Variant 2

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

• Resolution

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

  The given series converges by the comparison test, because we have bounded it above by the series \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2} \), which converges.

• Result

  The series converges.

• Variant 3

  \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2-4n+5} \).

• Resolution

  \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2-4n+5} \le \sum_{n=1}^{\infty}\frac{1}{\frac12 n^2} = 2 \sum_{n=1}^{\infty}\frac{1}{n^2} \).

• Result

  The series converges.

• Variant 4

  \(\displaystyle \sum_{n=1}^{\infty}\frac{2n^2+3n+4}{2n^2+5} \)

• Resolution

  \(\displaystyle \sum_{n=1}^{\infty}\frac{2n^2+3n+4}{2n^2+5} \ge \sum_{n=1}^{\infty}\frac{2n^2+5}{2n^2+5}=\sum_{n=1}^{\infty} 1 \)

  We can also use the fact that the sequence \(\displaystyle \frac{2n^2+3n+4}{2n^2+5} \) converges to 1, not to 0.

• Result

  The sequence diverges.

• Variant 5

  \(\displaystyle \sum_{n=1}^{\infty}\frac{2n^2+3n+4}{(2n^2+5)^2} \)

• Resolution

  \(\displaystyle \sum_{n=1}^{\infty}\frac{2n^2+3n+4}{(2n^2+5)^2} \le \sum_{n=1}^{\infty}\frac{9n^2}{4n^4} = \frac94 \sum_{n=1}^{\infty}\frac1{n^2} \)

• Result

  The sequence converges.

• Variant 6

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

• Resolution

  \(\displaystyle \sum\limits_{n=1}^{\infty} \frac{2n^2 + 1}{n^3} \ge \sum\limits_{n=1}^{\infty} \frac{2n^2}{n^3} \ge 2 \sum\limits_{n=1}^{\infty} \frac1n \).

• Result

  The sequence diverges.

• Variant 7

  \(\displaystyle \sum\limits_{n=1}^{\infty} \frac{n^3 - 1}{n^4 + n^2}. \)

• Resolution

  \(\displaystyle \sum\limits_{n=1}^{\infty} \frac{n^3 - 1}{n^4 + n^2} = \sum\limits_{n=2}^{\infty} \frac{n^3 - 1}{n^4 + n^2} \ge \sum\limits_{n=2}^{\infty} \frac{\frac12n^3}{n^4 + n^4} = \frac14 \sum\limits_{n=2}^{\infty} \frac1n \)

• Result

  The sequence diverges.

• Variant 8

  \(\displaystyle \sum\limits_{n=1}^{\infty} \frac{6^n + 7^n}{8^n - 2^n}. \)

• Resolution

  \(\displaystyle \sum\limits_{n=1}^{\infty} \frac{6^n + 7^n}{8^n - 2^n} \le \sum\limits_{n=1}^{\infty} \frac{7^n + 7^n}{\frac12 {\cdot}8^n} = 4 \sum\limits_{n=1}^{\infty} \left(\frac{7}{8}\right)^n \)

  In the first step we used the inequalities \(6^n \leq 7^n\) and \(2^n \leq \frac12 8^n\). The series converges by the comparison test, because we have bounded it above (up to a constant factor) by a convergent geometric series.

• Result

  The series converges.

• Variant 9

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

• Resolution

  \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{(n+1)\sqrt{n+2}}\le \sum_{n=1}^{\infty}\frac{1}{n\sqrt{n}}\le= \sum_{n=1}^{\infty}n^{-\frac32} \)

  The series converges, because the series \(\displaystyle \sum\limits_{n=1}^{\infty} n^{\alpha} \) converges for every \(\alpha < -1\).

• Result

  The series converges.

• Variant 10

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

• Resolution

  Except for the first two terms in the series, we can use the estimate

  \(\displaystyle \sum_{n=3}^{\infty}\frac{1}{\sqrt{2n+1}\sqrt{2n+3}}\ge \sum_{n=3}^{\infty}\frac{1}{\sqrt{3n}\sqrt{3n}}= \frac13\sum_{n=3}^{\infty}\frac1n \)

• Result

  The series diverges.

• Variant 11

  \(\displaystyle \sum\limits_{n=2}^{\infty} \frac{1}{\root 5 \of{n^4 + n} \root 3 \of{n-2}}. \)

• Resolution

  For \(n\ge 4\),

  \(\displaystyle \frac{1}{\root 5 \of{n^4 + n} \root 3 \of{n-2}} \leq \frac{1}{\root 5 \of{n^4} \root 3 \of{n-\frac{n}2}} = \frac{1}{\root 3 \of{\frac12} n^{\frac45} n^{\frac13}} = \frac{1}{\root 3 \of{\frac12} n^{\frac{17}{15}}} \)

  So we have the estimate \(\displaystyle \sum\limits_{n=4}^{\infty} \frac{1}{\root 5 \of{n^4 + n} \root 3 \of{n-2}} \le \sqrt[3]2\sum\limits_{n=4}^{\infty} n^{-\frac{17}{15}} \)

  (The first two terms do not influence the convergence of the series; their sum is finite).

• Result

  The series converges.