Collection of Maths Problems

Applications to the convergence of series

Task number: 2982

• Variant 1

  \(\displaystyle \sum_{n=1}^{\infty}\sin\left(\frac{1}{n}\right) \)

• Hint

  Use Heine's theorem:

  \(\displaystyle \lim_{x\to a}f(x)=A\text{ and }\lim_{n\to\infty}x_n=a, \text{ imply that }\lim_{n\to\infty}f(x_n)=A. \)

• Resolution

  From the known limit \(\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1 \) we know that when \(x\) is small, \(\sin x\) behaves roughly like \(x\). So we can guess that we can compare the series \(\sin \tfrac{1}{n}\) with the divergent series \(\tfrac{1}{n}\).

  From the relationships \(\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1 \) and \(\displaystyle \lim_{n\to\infty}\frac{1}{n}=0 \) we get \(\displaystyle \lim_{n\to\infty}\frac{\sin\tfrac{1}{n}}{\tfrac{1}{n}}=1 \).

  Then by the limit comparison test our series converges if and only if the series \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n} \) converges. But this series in fact diverges, so the given series diverges as well.

• Variant 2

  \(\displaystyle \sum_{n=1}^{\infty}\left(1-\cos\left(\frac{1}{n}\right)\right)n^{\alpha} \) depending on the parameter \(\alpha\in\mathbb R\).

• Resolution

  We use the fact that \(\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}. \)

  By Heine's theorem, \(\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}\text{ and } \lim_{n\to\infty}\frac{1}{n}=0 \text{ imply that } \lim_{n\to\infty}\frac{1-\cos\tfrac{1}{n}}{(\tfrac{1}{n})^2}=\frac{1}{2}. \)

  So if we let \(\displaystyle a_n=\left(1-\cos\left(\frac{1}{n}\right)\right)n^{\alpha}\), \(\displaystyle b_n=\frac{1}{n^2}n^{\alpha}, \) then we easily obtain \(\displaystyle \lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{1-\cos\tfrac{1}{n}}{(\tfrac{1}{n})^2} =\frac{1}{2}\in(0,\infty). \)

  By the limit comparison test, the given series converges if and only if the series \(\displaystyle \sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}n^{\alpha-2} \) converges. This series will converge when \(\alpha-2<-1\) or \(\alpha<1\).