Resolved: "Given series: ∑k=3∞[1k−1k+1] does this series converge or..."

Armorikam

Answered question

2021-02-20

Given series:

\sum_{k = 3}^{\infty} [\frac{1}{k} - \frac{1}{k + 1}]

does this series converge or diverge?
If the series converges, find the sum of the series.

Nathanael Webber

Skilled2021-02-21Added 117 answers

Given series is

\sum_{k = 3}^{\infty} (\frac{1}{k} - \frac{1}{k + 1})

A series

S = \sum_{k = 1}^{\infty} (a_{k})

is said to coverge only if the limit of its partial sum i.e.

lim_{n \to \infty} S_{n}

exist.
Here

S_{n} = \sum_{k = 1}^{\infty} (a_{k})

The sum of the series S is given by,

S = lim_{n \to \infty} S_{n}

The partial sum of the given series is,

S_{n} = \sum_{k = 3}^{\infty} (\frac{1}{k} - \frac{1}{k + 1})

Expand it

S_{n} = (\frac{1}{3} - \frac{1}{3 + 1}) + (\frac{1}{4} - \frac{1}{4 + 1}) + (\frac{1}{5} - \frac{1}{5 + 1}) + . . . + (\frac{1}{n} - \frac{1}{n + 1})

S_{n} = \frac{1}{3} - \frac{1}{n + 1}

Evaluate the limit

S = lim_{n \to \infty} S_{n}

S = lim_{n \to \infty} (\frac{1}{3} - \frac{1}{n + 1})

S = \frac{1}{3} - 0

S = \frac{1}{3}

Since the limit exist therefore the series converges and the sum of the series is given by,

\sum_{k = 3}^{\infty} (\frac{1}{k} - \frac{1}{k + 1}) = \frac{1}{3}

Answer
Series converges,

\sum_{k = 3}^{\infty} (\frac{1}{k} - \frac{1}{k + 1}) = \frac{1}{3}

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