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 $a-a_1$.