1. ## convergence

1st) I need to investigate convergence using convergence of integral feature
$\displaystyle \sum_{n=1}^{inf}\frac{1}{\sqrt{n}}$
From limit I see that it's 0, but how do I prove it using that method?

2nd) Here I need to investigate convergence using series required condition of convergence
$\displaystyle \sum_{n=1}^{\inf}\cos^2n$

2. The series...

$\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ (1)

... diverges and You can prove that considering that is...

$\displaystyle \displaystyle \sum_{n=1}^{N} \frac{1}{\sqrt{n}} > 1 + \int_{1}^{N} \frac{dx}{\sqrt{1+x}}$ (2)

But a primitive of the function under integral sign in (2) is $\displaystyle \displaystyle \frac{\sqrt{1+x}}{2}$ so that the integral in (2) diverges and the same is for the series (1)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Originally Posted by chisigma
But a primitive of the function under integral sign in (2) is $\displaystyle \displaystyle \frac{\sqrt{1+x}}{2}$
isn't primitive $\displaystyle 2\sqrt{x+1}+C$ ?
this one also diverge so proving doesn't change. Thanks

4. for the 1)st question you know that 1/$\displaystyle \sqrt{n}$ $\displaystyle \ge$ 1/n $\displaystyle \forall$ n $\displaystyle \in$ N (natural numbers). and you know that $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ is the harmonic series that diverges. so by comparison test $\displaystyle \sum_{n=1}^{\infty}\frac{1}\sqrt{n}$ diverges. since $\displaystyle \sum$1/n diverges $\displaystyle \forall$ c $\displaystyle \not=$ 0 c$\displaystyle \sum_{n=1}^{\infty}$1/n diverges. therefore the radius of convergence is 0.

2nd) there is a theorem that states that $\displaystyle \sum_{n=1}^{\infty}a_n$ converges $\displaystyle \Longrightarrow$ $\displaystyle \lim_{n\to\infty}a_n$= 0
which means if $\displaystyle \lim_{n\to\infty}a_n\not=$0 $\displaystyle \Longrightarrow$ $\displaystyle \sum_{n=1}^{\infty}a_n$ diverges. since 1$\displaystyle \ge$$\displaystyle \cos^2$n$\displaystyle \ge$0 $\displaystyle \forall$ n $\displaystyle \in$ N, for a given n you can find n$\displaystyle \prime\in$ N such that cos n$\displaystyle \prime\not=$0, which means $\displaystyle \cos^2$n$\displaystyle \prime\not=0$. therefore $\displaystyle \sum_{n=1}^{\infty}\cos^2{n}$ diverges. once again the radius of convergence is 0.
hope this is helpful.