Cfrgtkky

1. Investigate convergence of $sum_{n=2}^{infty} frac{1}{n^{ln n}}$.

$frac{1}{n^{ln n}}<frac{1}{n^2}, sum frac{1}{n^2}=pi/6$, so the original series converges.
(To verify the convergence of $sumfrac{1}{n^2}$, we can first apply the telescopic equivalence, and then use the ratio test).

2. Investigate convergence of $sum frac{2^ncdot (n!)^2}{4cdot 11 cdot cdots cdot (2n^2+n+1)}$.

Let's use Raabe's test.

$n bigg( frac{2^ncdot (n!)^2}{4cdot 11 cdot cdots cdot (2n^2+n+1)}cdot frac{4cdot 11 cdot cdots cdot (2n^2+n+1) cdot (2(n+1)^2+(n+1)+1)}{2^ncdot (n!)^2cdot 2 cdot (n+1)^2} -1 bigg)= n bigg( frac{1}{1}cdot frac{ 2(n+1)^2+(n+1)+1}{2 cdot (n+1)^2} -1 bigg) = n bigg( frac{1}{1}cdot frac{ 2(n+1)^2+(n+1)+1 - 2 cdot (n+1)^2}{2 cdot (n+1)^2} bigg) = n bigg( frac{ (n+1)+1}{2 cdot (n+1)^2} bigg) = frac{n^2 + o(n^2)}{2n^2 + o(n^2)}longrightarrow frac{1}{2}<1,$

so the series diverges.

3. Investigate convergence of $sum (1+frac{2n^2+n}{3n^3-nsin n})^{frac{n^2 + n cos n}{n+sin n}}$.

We know that, for a series to converge, it's necessary for its $a_n$ to tend to zero.

$(1+frac{2n^2+n}{3n^3-nsin n})^{frac{n^2 + n cos n}{n+sin n}}$

The fraction inside tends to zero (dividing the nominator and the denominator by $n^2$), so we look for the exponent:

$(1+frac{2n^2+n}{3n^3-nsin n})^{frac{n^2 + n cos n}{n+sin n} cdot frac{2n^2+n}{3n^3-nsin n} cdot frac{3n^3-nsin n}{2n^2+n} }$

$e^{{frac{n^2 + n cos n}{n+sin n} cdot frac{2n^2+n}{3n^3-nsin n}}}$

$e^{frac{2n^4 + o(n^4)}{3n^4+o(n^4)}}$

$e^{frac{2}{3}} neq 0$,

so the series diverges.

4. Investigate convergence of $sum frac{1}{(ln n)^{ln n}}$.

Convergence of $frac{1}{(ln n)^{ln n}} iff $ convergence of $frac{2^n}{(n ln 2)^{n ln 2}}$.

Applying Cauchy's test,

$limbigg( frac{2^n}{(n ln 2)^{n ln 2}} bigg)^{1/n}=limfrac{2}{(n ln 2)^{ ln 2}}=0<1$, so the series converges.

My intuition tells me that in terms of limits and convergence, $frac{2^n}{(n ln 2)^{n ln 2}}$ must be in some sense equivalent to the $frac{2^n}{n^{n}}$, which is converging big time (ratio test to see). I will be very grateful if someone could tell me how to formalize the idea, or if it is even correct at all.

5. Investigate convergence of $sum frac{2cdot 5 cdot cdots cdot (3n-4)}{3^n cdot n!}$.

Let's use Raabe's test.

$lim n bigg( frac{2cdot 5 cdot cdots cdot (3n-4)}{3^n cdot n!} cdot frac{3^n cdot n! cdot 3 cdot (n+1)}{2cdot 5 cdot cdots cdot (3n-4) cdot (3(n+1)-4)} - 1 bigg)$

$lim n bigg( frac{1}{1} cdot frac{ 3 cdot (n+1)}{3(n+1)-4} - 1 bigg)$

$lim n bigg( frac{3n+3-3n+1}{3n-1} bigg)$

$lim bigg( frac{4n}{3n-1} bigg)$

$frac{4}{3} > 1,$

so the series converges.

6. 
   $sum a_n$ is absolutely converging (i.e. $sum |a_n|$ converges), and $sum b_n$ is diverging. Does $sum a_n+b_n$ converge?

I am having a hard time with this one.

We know from the lectures that if $sum a_n$ is absolutely converging, then $sum a_n$ is converging.

From the definition, $forall epsilon>0 exists N_a : forall n, m>N_a Longrightarrow |a_n+cdots+a_m|<epsilon.$

Now, I'm not sure if I have written the negation correctly:

$exists epsilon>0 forall N_b exists n, m>N_b Longrightarrow |b_n+cdots+b_m|geepsilon.$

Taking the difference of one from another, I can get the desired (?)

$|a_n+cdots+a_m|-epsilon<epsilon-|b_n+cdots+b_m|$

$|a_n+cdots+a_m|+|b_n+cdots+b_m|<2epsilon,$

now, from the triangle inequality,

$|a_n+cdots+a_m+b_n+cdots+b_m|<2epsilon.$

(The series is diverging).

I feel like neglecting the quantors can land me in trouble.

7. With what range of values of $alpha>0$ does $sum frac{3^n}{(2n)!n^{alpha}}$ converge?

If we prove that with $alpha=0$ the series converges, then we have shown it converges for all the positive alphas. Ratio test tells us

$frac{3^ncdot 3}{(2n)! cdot (2n+1) cdot (2n+2)}cdot frac{(2n)!}{3^n}$

$frac{3}{(2n+1) cdot (2n+2)} longrightarrow 0<1,$

so the series converges.

Thank you very much. Alternative proofs are welcome.