Collection of Maths Problems

Using the definition

Task number: 2929

• Variant 1

  Show that the harmonic series \(\displaystyle \sum_{n=1}^{\infty}\frac1n \) diverges.

• Resolution

  The sequence of partial sums \(s_n\) satisfies

  \(s_{2n}-s_nk= \frac1{n+1}+\frac1{n+2}+…+\frac1{2n}\ge n\cdot\frac1{2n}=\frac12\).

  So for \(\varepsilon=\frac12\) and any \(n\) it is true that \(|s_{2n}-s_{n}|\ge \varepsilon\), so the sequence of partial sums does not satisfy the Bolzano-Cauchy criterion and therefore diverges.

• Variant 2

  Show that the series \(\displaystyle \sum_{n=1}^{\infty} \frac1{\sqrt n}\) diverges.

• Resolution

  \(s_{2n}-s_n= \frac1{\sqrt{n+1}}+\frac1{\sqrt{n+2}}+…+\frac1{\sqrt{2n}}\ge \frac{n}{\sqrt{2n}}=\frac{\sqrt{2n}}2\).

  So for \(\varepsilon=\frac{\sqrt{2}}2\) and any \(n\) we have \(|s_{2n}-s_{n}|\ge \varepsilon\), so the sequence of partial sums diverges.

• Variant 3

  Investigate the convergence or divergence of the series \(\displaystyle \sum_{n=1}^{\infty} \ln\left(1+\frac1n\right)\).

• Resolution

  \(\displaystyle s_n=\sum_{k=1}^n \ln\left(1+\frac1k\right)= \ln\left(\prod_{k=1}^n \left(1+\frac1k\right)\right)= \ln\left(\prod_{k=1}^n \frac{k+1}k\right)= \ln\left(\frac21\cdot\frac32\cdots\frac{n+1}n\right)= \ln({n+1}) \)

  And so \(\displaystyle \lim_{n\to \infty} s_n=\lim_{n\to \infty} \ln({n+1})= \infty \).

• Result

  The sequence diverges.

• Variant 4

  Let \(\displaystyle \lim_{n\to \infty} a_n = a \in \mathbb R\). Determine \(\displaystyle \sum_{n=1}^{\infty} (a_{n+1}-a_n) \).

• Resolution

  Consider the sequence of partial sums

  \(\displaystyle s_n=\sum_{k=1}^n (a_k+1-a_k)=(a_2-a_1)+(a_3-a_2)+(a_4-a_3)+…+(a_{n+1}-a_n)=a_{n+1}-a_1 \)

  And so \(\displaystyle \lim_{n\to \infty} s_n=\lim_{n\to \infty} a_{n+1}-a_1= a-a_1 \)

• Result

  The sum of the series is \(a-a_1\).