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Prove the convergence of

$$ \sum_{n = 1}^{\infty} {\sqrt{\, 2n - 1\,}\,\ln\left(4n + 1\right) \over n\left(n+1\right)} $$

I've been struggling for hours on this. By the textbook we have the limit test, comparison test, asymptotic test and integral test. When I use the asymptotic test my quotient ends up as $0$ or $\infty$, and I cannot prove anything. Since I am self studying, I have no one to turn to. The rest of the questions were easy, but I cant figure this out. Please help!

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3 Answers 3

$\ln n$ grows much, much, much more slowly than any positive power of $n$. So for $n$ large enough,

$$ \frac{\sqrt{2n-1}\ln(4n+2)}{n(n+1)} < \frac{\sqrt{2n}*n^{1/10}}{n^2}<\frac{2}{n^{14/10}}.$$

Use the Comparison Test (then the Integral Test). You may have to justify the first inequality, depending on who's asking.

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The two problems yield to basically the same technique. We look at the first one, which is somewhat harder.

First let us argue informally. Note that $\sqrt{2n-1}$ in the long run behaves like $\sqrt{2}\sqrt{n}$. Since we won't worry about constants, let us say it behaves like $\sqrt{n}$.

The bottom behaves like $n^2$, so $\frac{\sqrt{2n-1}}{n(n+1)}$ behaves like $\frac{1}{n^{3/2}}$.

The term $\ln(4n+1)$, in the long run, grows very slowly, more slowly than any power of $n$. So in the long run, in particular, it grows more slowly than $n^{1/4}$. Thus, in the long run, our original expression goes to $0$ faster than $\frac{1}{n^{5/4}}$.

But $\sum_1^\infty \frac{1}{n^{5/4}}$ converges. It follows that our original series converges.

In essence, in $\frac{1}{n^{3/2}}$, the exponent $3/2$ is plenty big enough to ensure convergence. So we grab an $n^{1/4}$ from it to "kill" the $\ln(4n+1)$ term. That leaves us with $\frac{1}{n^{5/4}}$, which still goes to $0$ fast enough.

Now it is time to turn the above into a more formal argument, say by using the Limit Comparison Test. Let $a_n$ be the $n$-th term of our series. We show that $$\lim_{n\to\infty} \frac{a_n}{\frac{1}{n^{5/4}}}=0.$$ Since $n^{5/4}=n^{3/2}/n^{1/4}$, we want to show that $$\lim_{n\to\infty} \frac{\sqrt{2n-1}\,n^{3/2}}{n(n+1)}\frac{\ln(4n+1)}{n^{1/4}}=0.$$ We can use L'Hospital's Rule to show that $\lim_{n\to\infty}\frac{\ln(4n+1)}{n^{1/4}}=0$. Standard techniques show that $$\lim_{n\to\infty}\frac{\sqrt{2n-1}\,n^{3/2}}{n(n+1)}=\sqrt{2}.$$ This completes the Limit Comparison argument.

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This technique also works. I think this is what the book was getting at. Thanks! –  Haresh Jan 4 at 6:12
    
You are welcome. You will see that exactly the same idea will work for your second problem. We need not have grabbed $n^{1/4}$ to kill the $\ln(4n+1)$ term. for example, $n^{0.1}$, or $n^{0.4}$, would have worked. But $n^{0.5}$ would not, since it takes away too much from $n^{3/2}$. –  André Nicolas Jan 4 at 6:18

I would suggest you start with some rough hand-wavy approximations and then make it rigorous. For example, when $n$ is large, $$\frac {\sqrt{2n-1}}{n(n+1)}\approx \frac {\sqrt {2n}}{n^2}=\sqrt 2 n^{-3/2}$$ and $\ln (4n+1) \sim \ln n$.

Let's try integrating $$\int x^{-3/2} \ln x \,dx.$$ Let $u = \ln x$, $du = \frac 1 x dx$, $dv = x^{-3/2}$, $v=-2 x^{-1/2}$. This gives an integral of \begin{align*} -2x^{-1/2}\ln x - \int -2x^{-3/2}\,dx &=-2x^{-1/2}\ln x-4x^{-1/2} +C\\ &=-2\frac{\ln x}{\sqrt x}-4\frac{1}{\sqrt x}+C. \end{align*} So the integral from, say, $1$ to $+\infty$ of this thing is quite clearly finite.

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I've tried that. Ended up with $\frac{{1}/{n^(1/2)}}$ which doesn't converge, while the textbook says it does. –  Haresh Jan 4 at 5:45
    
@Haresh, I think you made a mistake somewhere along the way. I suggest you try it again. What you end up with is the result of approximating $\ln n$ by something proportional to $n$, which would be a terrible approximation. –  dfeuer Jan 4 at 5:46
    
Okay, I think the I mixed up my work with my second question, which I removed to meet guidelines, and the problem is that the textbook does not tell us why $\frac{\ln n}{n^p}}$, for p greater than 1 converges. How would I prove this? Both of the problems boil down to this series. –  Haresh Jan 4 at 5:57
    
@Haresh, check out math.uconn.edu/~kconrad/blurbs/analysis/growth.pdf –  dfeuer Jan 4 at 6:02
    
Thanks so much, I now see that I had to use integration by parts. I would never have figured this out! –  Haresh Jan 4 at 6:10

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