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 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.
You can study the function $\displaystyle f_k$ deifned by $\displaystyle f_k\left(x\right) = \frac 1{\left|x+k\right|^a}$ ($\displaystyle x$ can't be an integer).
But it seem that it's not a bounded function: what is $\displaystyle \lim_{x\to -k}f_k\left(x\right)$
I don't think you can have such $\displaystyle M_k$ because $\displaystyle f_k(x)=\frac{1}{1+|x+k|^a} \leq 1$ but actually attains this vaue at $\displaystyle -k$ so the M-test can't be applied on the whole line (Maybe in some interval you could).
Edit: For example for $\displaystyle k\geq 1$ we have $\displaystyle f_k(x)\leq \frac{1}{k^a}$ on $\displaystyle (0,\infty)$ and $\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^a} <\infty$. Similarly you can apply the M-test to $\displaystyle f_k$ for $\displaystyle k\leq -1$ on $\displaystyle (-\infty , 0)$.
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?