Collection of Maths Problems

Series as limits

Task number: 2900

• Variant 1

  \(\displaystyle \lim_{n\to\infty}\left(\frac{\sum_{k=1}^n k}{n+2}-\frac{n}2\right) \).

• Resolution

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

• Result

  The limit equals \(-\frac12\).

• Variant 2

  \(\displaystyle \lim_{n\to\infty}\frac1{n^4}\sum_{k=1}^n k^3 \).

• Resolution

  \(\displaystyle \lim_{n\to\infty}\frac1{n^4}\sum_{k=1}^n k^3= \lim_{n\to\infty}\frac1{n^4}\left(\frac{n(n+1)}{2}\right)^2= \lim_{n\to\infty}\frac{1+2n^{-1}+n^{-2}}4=\frac14 \).

• Result

  The limit equals \(\frac14\).

• Variant 3

  \(\displaystyle \lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^n \lfloor xk\rfloor \) depending on the real parameter \(x\).

• Resolution

  By the squeeze lemma

  \(\displaystyle \lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^n \lfloor xk\rfloor \le \lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^n xk = \lim_{n\to\infty}\frac{x}{n^2} \cdot\frac{n(n+1)}2 = \frac{x}2 \)

  \(\displaystyle \lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^n \lfloor xk\rfloor \ge \lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^n (xk-1) = \lim_{n\to\infty}\left(\frac{x}{n^2} \cdot\frac{n(n+1)}2 - \frac1n\right) = \frac{x}2 \)

• Result

  The limit equals \(\frac x2\).

• Variant 4

  \(\displaystyle \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k(k+1)}. \)

• Resolution

  In view of the identity \(\displaystyle \frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1} \) we can transform the given expression to

  \(\displaystyle \sum_{k=1}^n\frac{1}{k(k+1)}=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right) +…+\left(\frac{1}{n}-\frac{1}{n+1}\right)=1-\frac{1}{n+1} \).

  And now we can easily determine that \(\displaystyle \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k(k+1)}=\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)=1. \)

• Result

  The limit equals \(1\).