Collection of Maths Problems

Series with parameters

Task number: 2935

• Variant 1

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

• Resolution

  \(\displaystyle \lim_{n\to\infty} \frac{1}{1+a^n}= \frac{1}{1+\lim\limits_{n\to\infty}a^n}= \begin{cases} \frac{1}{1+0}=1 & \text{ for } a\in\langle0{,}1),\\ \frac{1}{1+1}=\frac12 & \text{ for } a=1,\\ \frac{1}{1+\infty}=0 & \text{ for } a>1. \end{cases} \)

  So the series cannot converge for \(a\le1\).

  For \(a>1\) we can use the estimate \(\displaystyle \lim_{n\to\infty} \frac{1}{1+a^n}\le \lim_{n\to\infty} \left(\frac{1}{a}\right)^n \), which is an upper bound formed by a convergent geometric series.

• Result

  The series converges for \(a>1\).

• Variant 2

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

• Resolution

  \(\displaystyle \sum_{n=1}^{\infty} \frac{a^n}{1+a^n}= \sum_{n=1}^{\infty} \frac{1}{1+\left(\frac1a\right)^n}\)

  Now the result from the preceding variant gives us an answer directly.

• Result

  The series converges for \(a\in(0{,}1)\).

• Variant 3

  \(\displaystyle \sum_{n=1}^{\infty} \frac{a^n}{n!}\)

• Resolution

  By the ratio test:

  \(\displaystyle \lim_{n\to\infty}\frac{a_{n+1}}{a_n}= \lim_{n\to\infty}\frac{\frac{a^{n+1}}{(n+1)!}}{\frac{a^n}{n!}}= \lim_{n\to\infty}\frac{a}{n+1}=0 \)

• Result

  The series converges for all positive (and negative) \(a\).

• Variant 4

  \(\displaystyle \sum_{n=1}^{\infty}\frac{1+na}{\sqrt{n^2+n^6a}} \)

• Resolution

  For \(n>a\) we estimate:

  \(\displaystyle \sum_{n=1}^{\infty}\frac{1+na}{\sqrt{n^2+n^6a}}\le \sum_{n=1}^{\infty}\frac{2na}{\sqrt{n^6a}}= 2\sqrt{a}\sum_{n=1}^{\infty}n^{-2} \)

• Result

  The series converges for all positive values of \(a\) (and even for \(a\in(-1{,}0\rangle\)).

• Variant 5

  \(\displaystyle \sum_{n=1}^{\infty}n^4a^n \)

• Resolution

  For \(a\ge 1\) we have \(n^4a^n\ge1\), and so the series cannot converge.

  For \(a\in(0{,}1)\) we use the ratio test:

  \(\displaystyle \lim_{n\to\infty}\frac{a_{n+1}}{a_n}= \lim_{n\to\infty}\frac{(n+1)^4a^{n+1}}{n^4a^n}= a\left(\lim_{n\to\infty} 1+\frac1n\right)^4=a<1\).

• Result

  The series converges for \(a\in(0{,}1)\) (and even for \(a\in(-1{,}0\rangle\)).