Collection of Maths Problems

Series

Task number: 2901

• Variant 1

  \(\displaystyle \sum_{n=0}^{\infty}\frac{3^n-2^{n+1}}{6^n} \).

• Resolution

  \(\displaystyle \sum_{n=0}^{\infty}\frac{3^n-2^{n+1}}{6^n}= \sum_{n=0}^{\infty}\left(\frac36\right)^n- 2\sum_{n=0}^{\infty}\left(\frac26\right)^n= 2-2\cdot\frac32=-1 \).

  We use the fact that \(\displaystyle \sum_{n=0}^{\infty}q^n=\frac1{1-q}\) for \(q<1\).

• Result

  The sum of the sequence is \(-1\).

• Variant 2

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

• Resolution

  \(\displaystyle s_n=\sum_{k=2}^n\frac{1}{k^2-1}= \sum_{k=2}^n\frac{1}{(k-1)(k+1)}= \frac12\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k+1}\right)=\\ \frac12 \left[ \left(\frac{1}{1}-\frac{1}{3}\right)+ \left(\frac{1}{2}-\frac{1}{4}\right)+ \left(\frac{1}{3}-\frac{1}{5}\right)+ …+ \left(\frac{1}{n-1}-\frac{1}{n+1}\right) \right] = \frac12 \left[ \frac{3}{2}-\frac{1}{n}-\frac{1}{n+1}\right] \)

  \(\displaystyle \sum_{n=2}^\infty\frac{1}{n^2-1}= \lim_{n\to\infty}s_n= \lim_{n\to\infty}\frac12 \left[\frac{3}{2}-\frac{1}{n}-\frac{1}{n+1}\right] =\frac34 \)

• Result

  The sum of the sequence is \(\frac 34\).

• Variant 3

  \(\displaystyle \sum_{n=1}^{\infty}\frac1{4n^2-1} \).

• Resolution

  \(\displaystyle s_n=\sum_{k=1}^n\frac{1}{4k^2-1}= \sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}= \frac12\sum_{k=2}^n\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)=\\ \frac12 \left[ \left(\frac{1}{1}-\frac{1}{3}\right)+ \left(\frac{1}{3}-\frac{1}{5}\right)+ …+ \left(\frac{1}{n-1}-\frac{1}{n+1}\right) \right] = \frac12 \left[ 1-\frac{1}{n+1}\right] \)

  \(\displaystyle \sum_{n=1}^\infty\frac{1}{4n^2-1}= \lim_{n\to\infty}s_n= \lim_{n\to\infty}\frac12 \left[1-\frac{1}{n+1}\right] =\frac12 \)

• Result

  The sum of the sequence is \(\frac12\).

• Variant 4

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

• Resolution

  \(\displaystyle \sum_{n=1}^\infty\frac{1}{n(n+5)}= \frac15\sum_{n=1}^\infty\left(\frac1n-\frac{1}{n+5}\right)=\frac15\left(1+\frac 12+ \frac13+\frac14+\frac15\right) \)

• Result

  The sum of the sequence is \(\frac{137}{300}\).

• Variant 5

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

• Resolution

  We use the relationship

  \(\displaystyle \frac1{n+1}-\frac1{n+2}=\frac{1}{(n+1)(n+2)}\) a \(\displaystyle \frac1{n}-\frac1{n+3}=\frac{3}{n(n+3)}\).

  \(\displaystyle \sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)(n+3)}= \frac12 \sum_{n=1}^\infty \left[\frac13\left(\frac1n-\frac1{n+3}\right)-\left(\frac1{n+1}-\frac1{n+2}\right) \right]=\)
  \(\displaystyle =\frac12 \left[\frac13\left(1+\frac12+\frac13\right)-\frac12 \right]=\frac1{18} \)

• Result

  The sum of the sequence is \(\frac1{18}\).

• Variant 6

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

• Result

  3