Collection of Maths Problems

Comparison test - difference of roots

Task number: 2931

• Variant 1

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

• Hint

  Expand an expression of type \(A - B\) using the fraction \(\frac{A+B}{A+B}\).

• Resolution

  \(\displaystyle \sum_{n=1}^{\infty}\left(\sqrt{n^2+1}-\sqrt{n^2-1}\right)= \sum_{n=1}^{\infty}\frac{{n^2+1}-(n^2-1)}{\sqrt{n^2+1}+\sqrt{n^2-1}}\ge \sum_{n=1}^{\infty}\frac2{\sqrt{4 n^2}+\sqrt{n^2}}= \frac23\sum_{n=1}^{\infty}\frac1n \).

• Result

  The series diverges.

• Variant 2

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

• Resolution

  \(\displaystyle \sum_{n=1}^{\infty}\left(\sqrt{n^3+1}-\sqrt{n^3-1}\right)= \sum_{n=1}^{\infty}\frac{{n^3+1}-(n^3-1)}{\sqrt{n^3+1}+\sqrt{n^3-1}}\le \sum_{n=1}^{\infty}\frac2{\sqrt{n^3}}= 2\sum_{n=1}^{\infty}n^{-\frac32} \).

• Result

  The series converges.

• Variant 3

  \(\displaystyle \sum_{n=1}^{\infty}\frac1n\left(\sqrt{n^2+1}-\sqrt{n^2-1}\right) \).

• Resolution

  \(\displaystyle \sum_{n=1}^{\infty}\frac1n\left(\sqrt{n^2+1}-\sqrt{n^2-1}\right)= \sum_{n=1}^{\infty}\frac{{n^2+1}-(n^2-1)}{n\left(\sqrt{n^2+1}+\sqrt{n^2-1}\right)}\le \sum_{n=1}^{\infty}\frac2{n\cdot n}= 2\sum_{n=1}^{\infty}n^{-2}\).

• Result

  The series converges.

• Variant 4

  \(\displaystyle \sum\limits_{n=2}^{\infty} \left( \sqrt{n^3 + 1} - \sqrt{n^3 - 3} \right). \)

• Resolution

  \(\displaystyle \sum_{n=2}^{\infty}\left(\sqrt{n^3+1}-\sqrt{n^3-3}\right)= \sum_{n=2}^{\infty}\frac{{n^3+1}-(n^3-3)}{\sqrt{n^3+1}+\sqrt{n^3-3}}\le \sum_{n=2}^{\infty}\frac4{\sqrt{n^3}}= 4\sum_{n=2}^{\infty}n^{-\frac32} \).

• Result

  The series converges.