Best solutions to - "Need to prove the sequence: an=1+122+132+…+1n2 converges."

Jamarion Kerr Answered2022-01-22

Need to prove the sequence:

a_{n} = 1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots + \frac{1}{n^{2}}

converges.

Answer & Explanation

Gordon Stephens

Expert

2022-01-23Added 10 answers

Notice that

2 k^{2} \geq k (k + 1) \Rightarrow \frac{1}{k^{2}} \leq \frac{2}{k (k + 1)}

\sum_{k = 1}^{\infty} \frac{2}{k (k + 1)} = \frac{2}{1 \times 2} + \frac{2}{2 \times 3} + \frac{2}{3 \times 4} + \dots

\sum_{k = 1}^{\infty} \frac{2}{k (k + 1)} = 2 ((1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots)

\sum_{k = 1}^{\infty} \frac{2}{k (k + 1)} = 2 (1) = 2

Therefore

\sum_{k = 1}^{\infty} \frac{1}{k^{2}} \leq 2

Micheal Hensley

Expert

2022-01-24Added 10 answers

a_{n} = \frac{1}{n^{2}}

and because

a_{n} > 0

we have

| a_{n} | = a_{n}

First: check the necessary condition

lim_{n \to \infty} n a_{n} = lim_{n \to \infty} \frac{1}{n = 0}

Second: check DAlemberts ratio test

lim_{n \to + \infty} | \frac{a_{n + 1}}{a_{n}} | = lim_{n \to + \infty} \frac{a_{n + 1}}{a_{n}} = lim_{n \to + \infty} {(\frac{n}{n + 1})}^{2} = 1

Third: Because the answer of DAlemberts test is 1, you should use Raabes

RizerMix

Expert

2022-01-27Added 437 answers

This here should work with

n \geq 1

s_{n} = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{1}{1} + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + . . . + \frac{1}{n^{2}}

b_{n} = \sum_{n = 1}^{\infty} \frac{1}{2^{n - 1}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^{n - 1}}

b_{n}

is directly compared greater than

s_{n}

s_{n} < b_{n}

and

b_{n}

converges, because of its ratio test :

\frac{1}{2^{n - 1 + 1}} / \frac{1}{2^{n - 1}} = \frac{2^{n - 1}}{2^{n}} = \frac{1}{2} < 1

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