We need to find the sum of the given series. ∑n=4∞(1n+1−1n+2)

We can find the sum by expanding the given series. ∑n=4∞(1n+1−1n+2)=limk→∞∑n=4k(1n+1−1n+2) =limk→∞[(14+1−14+2)+(15+1−15+2)+(16+1−16+2)+...+(1k+1−1k+2)] =limk→∞[(15−16)+(16−17)+(17−18)+...+(1k+1−1k+2)] =limk→∞[(15+(−16+16)+(−17+17)+(−18+18)+...+(−1k+1+1k+1)−1k+2] =[15+(−16+16)+(−17+17)+(−18+18)+...+(−1k+1+1k+1)−1k+2] limk→∞[15+0+0+0+...+0−1k−2] limk→∞[15−1k+2] =15−1∞+2=15−0=15 Answer: ∑n=4∞(1n+1−1n+2)=15

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Ramsey

Answered question

2021-01-31

We need to find the sum of the given series.

\sum_{n = 4}^{\infty} (\frac{1}{n + 1} - \frac{1}{n + 2})

Answer & Explanation

Benedict

Skilled2021-02-01Added 108 answers

We can find the sum by expanding the given series.

\sum_{n = 4}^{\infty} (\frac{1}{n + 1} - \frac{1}{n + 2}) = lim_{k \to \infty} \sum_{n = 4}^{k} (\frac{1}{n + 1} - \frac{1}{n + 2})

= lim_{k \to \infty} [(\frac{1}{4 + 1} - \frac{1}{4 + 2}) + (\frac{1}{5 + 1} - \frac{1}{5 + 2}) + (\frac{1}{6 + 1} - \frac{1}{6 + 2}) + . . . + (\frac{1}{k + 1} - \frac{1}{k + 2})]

= lim_{k \to \infty} [(\frac{1}{5} - \frac{1}{6}) + (\frac{1}{6} - \frac{1}{7}) + (\frac{1}{7} - \frac{1}{8}) + . . . + (\frac{1}{k + 1} - \frac{1}{k + 2})]

= lim_{k \to \infty} [(\frac{1}{5} + (- \frac{1}{6} + \frac{1}{6}) + (- \frac{1}{7} + \frac{1}{7}) + (- \frac{1}{8} + \frac{1}{8}) + . . . + (- \frac{1}{k + 1} + \frac{1}{k + 1}) - \frac{1}{k + 2}]

= [\frac{1}{5} + (- \frac{1}{6} + \frac{1}{6}) + (- \frac{1}{7} + \frac{1}{7}) + (- \frac{1}{8} + \frac{1}{8}) + . . . + (- \frac{1}{k + 1} + \frac{1}{k + 1}) - \frac{1}{k + 2}]

lim_{k \to \infty} [\frac{1}{5} + 0 + 0 + 0 + . . . + 0 - \frac{1}{k - 2}]

lim_{k \to \infty} [\frac{1}{5} - \frac{1}{k + 2}]

= \frac{1}{5} - \frac{1}{\infty + 2} = \frac{1}{5} - 0 = \frac{1}{5}

Answer:

\sum_{n = 4}^{\infty} (\frac{1}{n + 1} - \frac{1}{n + 2}) = \frac{1}{5}

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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