"I am confused on the following series: ∑n=11n(n+1)=1" was answered by our community

derlingasmh

Answered question

2022-01-23

I am confused on the following series:

\sum_{n = 1} \frac{1}{n (n + 1)} = 1

jojoann325

Beginner2022-01-24Added 10 answers

First consider the finite series:

\sum_{n = 1}^{m} (\frac{1}{n} - \frac{1}{n + 1}) = 1 - \frac{12}{+} \frac{12}{-} \frac{13}{+} \frac{13}{-} \frac{14}{+} \dots

This sum is telescoping, since it collapses like a telescope.
Everything is left except for the first and last term. Now what's the limit as

m \to \infty

Jason Olsen

Beginner2022-01-25Added 14 answers

First consider the partial sum. Let

S_{m} = \sum_{n = 1}^{m} \frac{1}{n (n + 1)}

. We then have

S_{m} = \sum_{n = 1}^{m} (\frac{1}{n} - \frac{1}{n + 1})

= \sum_{n = 1}^{m} \frac{1}{n} - \sum_{n = 1}^{n} \frac{1}{n + 1}

= \sum_{n = 1}^{m} \frac{1}{n} - \sum_{n = 2}^{m + 1} \frac{1}{n}

= 1 + \sum_{n = 2}^{m} \frac{1}{n} - \sum_{n = 2}^{m} \frac{1}{n} - \frac{1}{m + 1} = 1 - \frac{1}{m + 1}

Now

\sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} = lim_{m \to \infty} S_{m} = lim_{m \to \infty} (1 - \frac{1}{m + 1}) = 1

RizerMix

Expert2022-01-27Added 656 answers

This is an example of whats

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