- #1

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[tex]lim_{n\rightarrow\infty}n(s_n-s_{n-1})=\infty[/tex]

I've already gone pretty far afield in my hunt for such a sequence, so I thought I'd enlist the help of you fine folks in my search.

- Thread starter Poopsilon
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- #1

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[tex]lim_{n\rightarrow\infty}n(s_n-s_{n-1})=\infty[/tex]

I've already gone pretty far afield in my hunt for such a sequence, so I thought I'd enlist the help of you fine folks in my search.

- #2

mathman

Science Advisor

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If you are willing to accept conditionally convergent series, try s_{n}=(-1)^{n}/√n.

- #3

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[tex]lim_{n\rightarrow\infty}n(s_n-s_{n-1})=\infty[/tex]

I've already gone pretty far afield in my hunt for such a sequence, so I thought I'd enlist the help of you fine folks in my search.

He is talking about a sequence, not a series. At least that's what he said although he did use notation that suggests partial sums of a series. Your sequence converges (not conditionally). If he meant what he said, your suggestion may work depending on what he really wants. If n is even, n(sIf you are willing to accept conditionally convergent series, try s_{n}=(-1)^{n}/√n.

- #4

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I think mathman may have misspoke when he said series, his sequence will work, all I needed was the absolute value of n(s_n-s_n-1) to increase without bound. Although for my ultimate purpose it turns out to be insufficient.

I'm taking a look at Cesaro summation: [tex]\sigma_n=\frac{s_0+s_1+...+s_n}{n+1}[/tex]

and trying to find a case where [tex]\lim\sigma_n=\sigma[/tex] while [tex]\lim s_n\neq\sigma.[/tex]

And if I am interpreting my book correctly the condition for the sequence s_n given in my original post is a necessary condition for this to occur, but upon doing the computation with mathman's example it appears not sufficient:

[tex]Set\;\; s_0=0,\;\; define\;\; s_n=\sum_{k=1}^{n}\frac{(-1)^k}{\sqrt{k}}\;\; for\;\; n\geq 1,\;\; and\;\; assume\;\; that\;\; \sigma_n=\sigma.\;\;\;\; Then\;\; [/tex][tex]\sigma_n=\frac{1}{n+1}\sum_{i=1}^{n}\frac{(-1)^i(n-i+1)}{\sqrt{i}}[/tex][tex]=\frac{1}{n+1}\sum_{i=1}^{n}\frac{(-1)^i(n+1)}{\sqrt{i}}-\frac{1}{n+1}\sum_{i=1}^{n}\frac{(-1)^ii}{\sqrt{i}}[/tex][tex]=\sum_{i=1}^{n}\frac{(-1)^i}{\sqrt{i}}-\frac{1}{n+1}\sum_{i=1}^{n}\frac{(-1)^ii}{\sqrt{i}}[/tex][tex]=s_n+\frac{1}{n+1}\sum_{i=1}^{n}(-1)^i\sqrt{i}\rightarrow \lim s_n=\sigma\;\; as\;\; n\rightarrow\infty.[/tex]

Where the penultimate equality follows by definition, and the ultimate by our assumption.

So I'm curious if either of you can come up with a sequence which does provide different limits for sigma_n and s_n.

P.S. does anyone know how to not automatically go down to the next line every time one codes in some latex, it used to not do that but I think physicsforum like changed their latex format and now it does, thanks.

Edit: Actually I'm not entirely sure that [tex]\lim_{n\rightarrow\infty}\frac{1}{n+1}\sum_{i=1}^{n}(-1)^i\sqrt{i}=0.[/tex]

If not than mathman's sequence does work after all.

I'm taking a look at Cesaro summation: [tex]\sigma_n=\frac{s_0+s_1+...+s_n}{n+1}[/tex]

and trying to find a case where [tex]\lim\sigma_n=\sigma[/tex] while [tex]\lim s_n\neq\sigma.[/tex]

And if I am interpreting my book correctly the condition for the sequence s_n given in my original post is a necessary condition for this to occur, but upon doing the computation with mathman's example it appears not sufficient:

[tex]Set\;\; s_0=0,\;\; define\;\; s_n=\sum_{k=1}^{n}\frac{(-1)^k}{\sqrt{k}}\;\; for\;\; n\geq 1,\;\; and\;\; assume\;\; that\;\; \sigma_n=\sigma.\;\;\;\; Then\;\; [/tex][tex]\sigma_n=\frac{1}{n+1}\sum_{i=1}^{n}\frac{(-1)^i(n-i+1)}{\sqrt{i}}[/tex][tex]=\frac{1}{n+1}\sum_{i=1}^{n}\frac{(-1)^i(n+1)}{\sqrt{i}}-\frac{1}{n+1}\sum_{i=1}^{n}\frac{(-1)^ii}{\sqrt{i}}[/tex][tex]=\sum_{i=1}^{n}\frac{(-1)^i}{\sqrt{i}}-\frac{1}{n+1}\sum_{i=1}^{n}\frac{(-1)^ii}{\sqrt{i}}[/tex][tex]=s_n+\frac{1}{n+1}\sum_{i=1}^{n}(-1)^i\sqrt{i}\rightarrow \lim s_n=\sigma\;\; as\;\; n\rightarrow\infty.[/tex]

Where the penultimate equality follows by definition, and the ultimate by our assumption.

So I'm curious if either of you can come up with a sequence which does provide different limits for sigma_n and s_n.

P.S. does anyone know how to not automatically go down to the next line every time one codes in some latex, it used to not do that but I think physicsforum like changed their latex format and now it does, thanks.

Edit: Actually I'm not entirely sure that [tex]\lim_{n\rightarrow\infty}\frac{1}{n+1}\sum_{i=1}^{n}(-1)^i\sqrt{i}=0.[/tex]

If not than mathman's sequence does work after all.

Last edited:

- #5

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I assume you are aware that Cesaro summation is convergence preserving. Look at the sequence {aI'm taking a look at Cesaro summation: [tex]\sigma_n=\frac{s_0+s_1+...+s_n}{n+1}[/tex]

and trying to find a case where [tex]\lim\sigma_n=\sigma[/tex] while [tex]\lim s_n\neq\sigma.[/tex]

- #6

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- #7

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Yes. If a sequence converges then its Cesaro sum converges to the same limit. There is a general theorem by Toeplitz giving a conditions on a summability matrix that make it convergence preserving, which the Cesaro sum matrix satisfies. You can read about it at

http://en.wikipedia.org/wiki/Silverman–Toeplitz_theorem

It is also not difficult to prove that result for Cesaro summation directly.

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