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Question Number 161504 by mathocean1 last updated on 18/Dec/21

M o n t r e r \overset{`}{a} p a r t i r d u c r i t \overset{`}{e r e} d e

C a u c h y q u e U_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{k^{2}} e s t u n e

d e C a u c h y .

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S h o w b y u s i n g {C a u c h y}^{'} s s e q u e n c e

d e f i n i t i o n t h a t U_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{k^{2}} i s a

s e q u e n c e o f C a u c h y .

Answered by mindispower last updated on 20/Dec/21

\Leftrightarrow \forall ϵ > 0 \exists N \forall (n, m) ⩾ N ∣ U_{n} - U_{m} ∣< ϵ

s o i t ϵ > 0

U_{n} - U_{m} = \underset{k = m + 1}{\sum^{n}} \frac{1}{k^{2}} = S_{n, m}

f (x) = \frac{1}{x^{2}}

\frac{1}{{(k + 1)}^{2}} ⩽ \int_{k}^{k + 1} \frac{1}{x^{2}} d x ⩽ \frac{1}{k^{2}}

\Rightarrow \underset{k = m}{\sum^{n - 1}} \frac{1}{{(k + 1)}^{2}} < \frac{1}{k} - \frac{1}{k + 1}

S_{n, m} < \frac{1}{m} - \frac{1}{n} = \frac{n - m}{n m}

s o i t N \in N t e l l ? Q u e \frac{1}{N} < ϵ \Rightarrow N = [ϵ] + 1

\Rightarrow S_{n, m} = \frac{1}{m} - \frac{1}{n} < \frac{1}{N} < ϵ, \forall (n, m) > N

\Rightarrow U_{n} e s t d e c a u c h y

Commented by mathocean1 last updated on 21/Dec/21

T h a n k s s i r!