How to check convergence of sequence in complete metric space.
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Let ${x_n}$ and ${y_n}$ be two sequences in a complete metric space $(X,d)$ such that
$d(x_n,x_{n+1})lefrac{1}{n^2}$
$d(y_n,y_{n+1})le frac{1}{n}$ , for all $nin mathbb{N}.$
Then which sequence would converge? Justify.
Now since the space is complete so every Cauchy sequence is convergent. Let $X=mathbb{R}$ and $d$, the usual metric. So I take the sequence of partial sums of Harmonic series ${H_n}_{nge1}$, then
$$d(H_n,H_{n+1})=|H_{n+1}-H_n|=1/(n+1)le1/n$$
But $H_n$ is not a Cauchy sequence so it does not converge in $X$.
So 2 need not converge.
Is my reasoning correct? Also I have a feeling that 1 would always converge, but I haven't be able to prove it. Can anyone help me with that? Thanks.
sequences-and-series convergence metric-spaces cauchy-sequences
$endgroup$
add a comment |
$begingroup$
Let ${x_n}$ and ${y_n}$ be two sequences in a complete metric space $(X,d)$ such that
$d(x_n,x_{n+1})lefrac{1}{n^2}$
$d(y_n,y_{n+1})le frac{1}{n}$ , for all $nin mathbb{N}.$
Then which sequence would converge? Justify.
Now since the space is complete so every Cauchy sequence is convergent. Let $X=mathbb{R}$ and $d$, the usual metric. So I take the sequence of partial sums of Harmonic series ${H_n}_{nge1}$, then
$$d(H_n,H_{n+1})=|H_{n+1}-H_n|=1/(n+1)le1/n$$
But $H_n$ is not a Cauchy sequence so it does not converge in $X$.
So 2 need not converge.
Is my reasoning correct? Also I have a feeling that 1 would always converge, but I haven't be able to prove it. Can anyone help me with that? Thanks.
sequences-and-series convergence metric-spaces cauchy-sequences
$endgroup$
$begingroup$
for large $M < N,$ how big is $sum_{j =M+1}^N ; frac{1}{j^2} ; ; ? $
$endgroup$
– Will Jagy
Jan 30 at 2:17
add a comment |
$begingroup$
Let ${x_n}$ and ${y_n}$ be two sequences in a complete metric space $(X,d)$ such that
$d(x_n,x_{n+1})lefrac{1}{n^2}$
$d(y_n,y_{n+1})le frac{1}{n}$ , for all $nin mathbb{N}.$
Then which sequence would converge? Justify.
Now since the space is complete so every Cauchy sequence is convergent. Let $X=mathbb{R}$ and $d$, the usual metric. So I take the sequence of partial sums of Harmonic series ${H_n}_{nge1}$, then
$$d(H_n,H_{n+1})=|H_{n+1}-H_n|=1/(n+1)le1/n$$
But $H_n$ is not a Cauchy sequence so it does not converge in $X$.
So 2 need not converge.
Is my reasoning correct? Also I have a feeling that 1 would always converge, but I haven't be able to prove it. Can anyone help me with that? Thanks.
sequences-and-series convergence metric-spaces cauchy-sequences
$endgroup$
Let ${x_n}$ and ${y_n}$ be two sequences in a complete metric space $(X,d)$ such that
$d(x_n,x_{n+1})lefrac{1}{n^2}$
$d(y_n,y_{n+1})le frac{1}{n}$ , for all $nin mathbb{N}.$
Then which sequence would converge? Justify.
Now since the space is complete so every Cauchy sequence is convergent. Let $X=mathbb{R}$ and $d$, the usual metric. So I take the sequence of partial sums of Harmonic series ${H_n}_{nge1}$, then
$$d(H_n,H_{n+1})=|H_{n+1}-H_n|=1/(n+1)le1/n$$
But $H_n$ is not a Cauchy sequence so it does not converge in $X$.
So 2 need not converge.
Is my reasoning correct? Also I have a feeling that 1 would always converge, but I haven't be able to prove it. Can anyone help me with that? Thanks.
sequences-and-series convergence metric-spaces cauchy-sequences
sequences-and-series convergence metric-spaces cauchy-sequences
asked Jan 30 at 2:02


Kushal BhuyanKushal Bhuyan
5,08021246
5,08021246
$begingroup$
for large $M < N,$ how big is $sum_{j =M+1}^N ; frac{1}{j^2} ; ; ? $
$endgroup$
– Will Jagy
Jan 30 at 2:17
add a comment |
$begingroup$
for large $M < N,$ how big is $sum_{j =M+1}^N ; frac{1}{j^2} ; ; ? $
$endgroup$
– Will Jagy
Jan 30 at 2:17
$begingroup$
for large $M < N,$ how big is $sum_{j =M+1}^N ; frac{1}{j^2} ; ; ? $
$endgroup$
– Will Jagy
Jan 30 at 2:17
$begingroup$
for large $M < N,$ how big is $sum_{j =M+1}^N ; frac{1}{j^2} ; ; ? $
$endgroup$
– Will Jagy
Jan 30 at 2:17
add a comment |
1 Answer
1
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$begingroup$
Your reasoning is good.
For 2, use the triangle inequality to show that, for $m < n$,
$$d(x_n, x_m) le sum_{k = m}^{n - 1} d(x_k, x_{k+1}).$$
Cauchiness follows from the convergence of $sum frac{1}{n^2}$.
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1 Answer
1
active
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1 Answer
1
active
oldest
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$begingroup$
Your reasoning is good.
For 2, use the triangle inequality to show that, for $m < n$,
$$d(x_n, x_m) le sum_{k = m}^{n - 1} d(x_k, x_{k+1}).$$
Cauchiness follows from the convergence of $sum frac{1}{n^2}$.
$endgroup$
add a comment |
$begingroup$
Your reasoning is good.
For 2, use the triangle inequality to show that, for $m < n$,
$$d(x_n, x_m) le sum_{k = m}^{n - 1} d(x_k, x_{k+1}).$$
Cauchiness follows from the convergence of $sum frac{1}{n^2}$.
$endgroup$
add a comment |
$begingroup$
Your reasoning is good.
For 2, use the triangle inequality to show that, for $m < n$,
$$d(x_n, x_m) le sum_{k = m}^{n - 1} d(x_k, x_{k+1}).$$
Cauchiness follows from the convergence of $sum frac{1}{n^2}$.
$endgroup$
Your reasoning is good.
For 2, use the triangle inequality to show that, for $m < n$,
$$d(x_n, x_m) le sum_{k = m}^{n - 1} d(x_k, x_{k+1}).$$
Cauchiness follows from the convergence of $sum frac{1}{n^2}$.
answered Jan 30 at 2:19
Theo BenditTheo Bendit
20.1k12354
20.1k12354
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$begingroup$
for large $M < N,$ how big is $sum_{j =M+1}^N ; frac{1}{j^2} ; ; ? $
$endgroup$
– Will Jagy
Jan 30 at 2:17