- 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 $\sum\frac{1}{n^2}$, we can first apply the telescopic equivalence, and then use the ratio test).
- Investigate convergence of $\sum \frac{2^n\cdot (n!)^2}{4\cdot 11 \cdot \cdots \cdot (2n^2+n+1)}$.
Let's use Raabe's test.
$n \bigg( \frac{2^n\cdot (n!)^2}{4\cdot 11 \cdot \cdots \cdot (2n^2+n+1)}\cdot \frac{4\cdot 11 \cdot \cdots \cdot (2n^2+n+1) \cdot (2(n+1)^2+(n+1)+1)}{2^n\cdot (n!)^2\cdot 2 \cdot (n+1)^2} -1 \bigg)= n \bigg( \frac{1}{1}\cdot \frac{ 2(n+1)^2+(n+1)+1}{2 \cdot (n+1)^2} -1 \bigg) = n \bigg( \frac{1}{1}\cdot \frac{ 2(n+1)^2+(n+1)+1 - 2 \cdot (n+1)^2}{2 \cdot (n+1)^2} \bigg) = n \bigg( \frac{ (n+1)+1}{2 \cdot (n+1)^2} \bigg) = \frac{n^2 + o(n^2)}{2n^2 + o(n^2)}\longrightarrow \frac{1}{2}<1,$
so the series diverges.
- Investigate convergence of $\sum (1+\frac{2n^2+n}{3n^3-n\sin n})^{\frac{n^2 + n \cos n}{n+\sin n}}$.
We know that, for a series to converge, it's necessary for its $a_n$ to tend to zero.
$(1+\frac{2n^2+n}{3n^3-n\sin n})^{\frac{n^2 + n \cos n}{n+\sin n}}$
The fraction inside tends to zero (dividing the nominator and the denominator by $n^2$), so we look for the exponent:
$(1+\frac{2n^2+n}{3n^3-n\sin n})^{\frac{n^2 + n \cos n}{n+\sin n} \cdot \frac{2n^2+n}{3n^3-n\sin n} \cdot \frac{3n^3-n\sin n}{2n^2+n} }$
$e^{{\frac{n^2 + n \cos n}{n+\sin n} \cdot \frac{2n^2+n}{3n^3-n\sin n}}}$
$e^{\frac{2n^4 + o(n^4)}{3n^4+o(n^4)}}$
$e^{\frac{2}{3}} \neq 0$,
so the series diverges.
- 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,
$\lim\bigg( \frac{2^n}{(n \ln 2)^{n \ln 2}} \bigg)^{1/n}=\lim\frac{2}{(n \ln 2)^{ \ln 2}}=0<1$, 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.
- Investigate convergence of $\sum \frac{2\cdot 5 \cdot \cdots \cdot (3n-4)}{3^n \cdot n!}$.
Let's use Raabe's test.
$\lim n \bigg( \frac{2\cdot 5 \cdot \cdots \cdot (3n-4)}{3^n \cdot n!} \cdot \frac{3^n \cdot n! \cdot 3 \cdot (n+1)}{2\cdot 5 \cdot \cdots \cdot (3n-4) \cdot (3(n+1)-4)} - 1 \bigg)$
$\lim n \bigg( \frac{1}{1} \cdot \frac{ 3 \cdot (n+1)}{3(n+1)-4} - 1 \bigg)$
$\lim n \bigg( \frac{3n+3-3n+1}{3n-1} \bigg)$
$\lim \bigg( \frac{4n}{3n-1} \bigg)$
$\frac{4}{3} > 1,$
so the series converges.
- $\sum a_n$ 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:
$\exists \epsilon>0 \forall N_b \exists n, m>N_b \Longrightarrow |b_n+\cdots+b_m|\ge\epsilon.$
Taking the difference of one from another, I can get the desired (?)
$|a_n+\cdots+a_m|-\epsilon<\epsilon-|b_n+\cdots+b_m|$
$|a_n+\cdots+a_m|+|b_n+\cdots+b_m|<2\epsilon,$
now, from the triangle inequality,
$|a_n+\cdots+a_m+b_n+\cdots+b_m|<2\epsilon.$
(The series is diverging).
I feel like neglecting the quantors can land me in trouble.
- 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
$\frac{3^n\cdot 3}{(2n)! \cdot (2n+1) \cdot (2n+2)}\cdot \frac{(2n)!}{3^n}$
$\frac{3}{(2n+1) \cdot (2n+2)} \longrightarrow 0<1,$
so the series converges.
Thank you very much. Alternative proofs are welcome.