convergent series..

**minesweeper26** · January 29th 2010, 08:04 AM

i have a sequence of function and i cannot find the correct bound for it. can someone help me out here? thanks in advanced!

Let a>1; x any real number..

1/(1+|x+k|)^a < M(sub)k such that the series M(sub)k, k= -infinity to +infinity is convergent.

**girdav** · January 29th 2010, 10:20 AM

You can study the function $f_k$ deifned by $f_k\left(x\right) = \frac 1{\left|x+k\right|^a}$ ( $x$ can't be an integer).
But it seem that it's not a bounded function: what is $\lim_{x\to -k}f_k\left(x\right)$

**minesweeper26** · January 30th 2010, 07:04 PM

thanks for the reply.

i actually want M(sub)k to be a sequence of real numbers, so it must be independent of x.

**Jose27** · January 30th 2010, 07:35 PM

Originally Posted by minesweeper26

i have a sequence of function and i cannot find the correct bound for it. can someone help me out here? thanks in advanced!

Let a>1; x any real number..

1/(1+|x+k|)^a < M(sub)k such that the series M(sub)k, k= -infinity to +infinity is convergent.

I don't think you can have such $M_k$ because $f_k(x)=\frac{1}{1+|x+k|^a} \leq 1$ but actually attains this vaue at $-k$ so the M-test can't be applied on the whole line (Maybe in some interval you could).

Edit: For example for $k\geq 1$ we have $f_k(x)\leq \frac{1}{k^a}$ on $(0,\infty)$ and $\sum_{k=1}^{\infty} \frac{1}{k^a} <\infty$ . Similarly you can apply the M-test to $f_k$ for $k\leq -1$ on $(-\infty , 0)$ .

**minesweeper26** · January 30th 2010, 09:11 PM

Originally Posted by Jose27

I don't think you can have such $M_k$ because $f_k(x)=\frac{1}{1+|x+k|^a} \leq 1$ but actually attains this vaue at $-k$ so the M-test can't be applied on the whole line (Maybe in some interval you could).

Edit: For example for $k\geq 1$ we have $f_k(x)\leq \frac{1}{k^a}$ on $(0,\infty)$ and $\sum_{k=1}^{\infty} \frac{1}{k^a} <\infty$ . Similarly you can apply the M-test to $f_k$ for $k\leq -1$ on $(-\infty , 0)$ .

inequality is not true for all x.. note that x can have a negative value.. hence, f(sub)k may not be less than 1/k^a..

another thing, a is the exponent of the whole denominator not just the absolute value..

thanks anyway..

EDIT: i mean, how will you bound f(sub)k on (-inf,0) for k>0?

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