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 Analysis: $\displaystyle{ \lim_{x \to \infty} \frac{x}{\ln x} \sum_{n=0}^{\infty} \frac{1}{a^n+x}\; (a >1)}$
03-08-2010, 08:27 AM
Post: #1
 elim Moderator Posts: 581 Joined: Feb 2010 Reputation: 0
Analysis: $\displaystyle{ \lim_{x \to \infty} \frac{x}{\ln x} \sum_{n=0}^{\infty} \frac{1}{a^n+x}\; (a >1)}$
Determine whether it's convergent. if so, find the limit.
03-11-2010, 11:07 PM
Post: #2
 elim Moderator Posts: 581 Joined: Feb 2010 Reputation: 0
Solution
Note that $$\quad\small{\displaystyle{\sum_{n = 0}^{\infty}\frac {1}{a^n + x} = \int_{0}^{\infty} \frac {dt}{a^{[t]} + x}}\quad}$$ thus

$$\small{\displaystyle{\int_{0}^{\infty} \frac {dt}{a^{t} + x} < \sum_{n = 0}^{\infty}\frac {1}{a^n + x} < \int_{0}^{\infty} \frac {dt}{a^{t - 1} + x} = \int_{ - 1}^{\infty} \frac {dt}{a^{t} + x} < \int_{0}^{\infty} \frac {dt}{a^{t} + x} + 1/x}}$$.

Hence $$\quad\small{\displaystyle{\sum_{n = 0}^{\infty}\frac {1}{a^n + x} = \int_{0}^{\infty} \frac {dt}{a^{t} + x} + O(1/x)=\frac{\ln (1+x)}{x\ln a} + O(1/x) }}$$,

and so $$\;\;\small{\displaystyle{\lim_{x\to \infty} \frac{x}{\ln x}\sum_{n=0}^{\infty}\frac{1}{a^n+x}=\lim_{x\to \infty} \frac{x}{\ln x} \left(\frac{\ln (1+x)}{x\ln a} + O(1/x)\right)=\frac{1}{\ln a}}}.\quad\square$$
09-04-2012, 07:55 PM
Post: #3
 elim Moderator Posts: 581 Joined: Feb 2010 Reputation: 0
Related Problem
What's the relationship between the limit in original problem and $\;\;\small{\displaystyle{\lim_{x \to \infty} \sum_{n=0}^{\infty} \frac{a^n x}{(a^n +x)^2}}}\;$?

12-27-2012, 12:17 PM
Post: #4
 elim Moderator Posts: 581 Joined: Feb 2010 Reputation: 0
RE: Related Problem
Note that $\displaystyle{\frac{x}{\ln x}\sum_{n=0}^{\infty} \frac{1}{a^n + x} = \frac{\sum_{n=0}^{\infty} \frac{x}{a^n + x}}{\ln x}}$, and $\displaystyle{\frac{\left(\frac{x}{a^n + x} \right )'}{(\ln x)'} = \frac{a^n x}{(a^n + x)^2}}$

Fix $a > 1$, for $x \ge 0$, since $\displaystyle{\frac{1}{a^n + x} \le a^{-n},\; g(x) = \sum_{n = 0}^{\infty} \frac{1}{a^n + x}}\;$ converges uniformly

Moreover, since $\left| \left((a^n + x)^{-1} \right)' \right| = |- (a^n + x)^{-2}| \le a^{-2n}$, thus $\displaystyle{g'(x) = - \sum_{n=1}^{\infty} \frac{1}{(a^n + x)^2}}$

We then have $\displaystyle{\quad \left(\sum_{n = 0}^{\infty} \frac{x}{a^n + x} \right )' = (x g(x))' = g(x) + xg'(x) = \sum_{n = 0}^{\infty} \frac{a^n}{(a_n + x)^2}}$
Therefore $\displaystyle{ \sum_{n = 0}^{\infty} \frac{a^n x}{(a^n + x)^2} = \frac{\left(\sum_{n=0}^{\infty} \frac{x}{a^n + x} \right )'}{(\ln x)'}}$ which shows the relationship between two problems.

But what we can do with $\displaystyle{ \sum_{n = 0}^{\infty} \frac{a^n x}{(a^n + x)^2}}$ in order to find out $\displaystyle{\lim_{x\to +\infty} \sum_{n = 0}^{\infty} \frac{a^n x}{(a^n + x)^2}}$?
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