Question: "Write a telescopic series including trigonometric functions and..."

Susan Munoz

Answered question

2021-11-18

Write a telescopic series including trigonometric functions and write anothrt telescopic series including logarithmic functions. show how both are convergent.

Answer & Explanation

Egreane61

Beginner2021-11-19Added 16 answers

Step 1: Find a series using trigonometric functions.

\sum_{n = 1}^{\propto} (\sin \frac{1}{n} - \sin \frac{1}{n + 1}) = (\sin 1 - \sin \frac{1}{2}) + (\sin \frac{1}{2} - \sin \frac{1}{3}) + (\sin \frac{1}{3} - \sin \frac{1}{4}) + \dots . + (\sin \frac{1}{n} - \sin \frac{1}{n + 1}) + \dots

which is clearly a telescopic series.
Step 2: Check its convergence.
Now,

lim_{n \to\propto} (\sin 1 - \sin \frac{1}{n + 1}) = \sin 1

hence the series converges to

\sin 1

Step 3: Find a series using logarithmic functions.
Consider the series

\sum_{n = 1}^{\propto} \ln (\frac{n (n + 2)}{{(n + 1)}^{2}}) = \sum_{n = 1}^{\propto} \ln \frac{\frac{n}{n + 1}}{\frac{n + 1}{n + 2}} = \sum_{n = 1}^{\propto} [\ln (\frac{n}{n + 1}) - \ln (\frac{n + 1}{n + 2})]

which is a telescoping series as

(\ln \frac{1}{2} - \ln \frac{2}{3}) + (\ln \frac{2}{3} - \ln \frac{3}{4}) + \dots

Step 4: Check its convergence.
Now,

\sum_{n = 1}^{\propto} [\ln (\frac{n}{n + 1}) - \ln (\frac{n + 1}{n + 2})] = lim_{n \to\propto} (\ln \frac{1}{2} - \ln \frac{n + 1}{n + 2})

= lim_{n \to\propto} (\ln \frac{1}{2} - \ln \frac{1 + \frac{1}{n}}{1 + \frac{2}{n}}) = \ln \frac{1}{2} = - \ln 2

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