Is $\sum_{n=1}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$ convergent? Is it absolute or conditional convergence?
Hello I've come across this problem my teacher gave me. Consider the infinite series $\sum_{n=1}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$. It's obvious that when $n = 1$, the value doesn't exist. But I asked my teacher and she said we can separate the $\frac{1}{\ln(1)}+\sum_{n=2}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$, and since $\sum_{n=2}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$ is conditional convergent, then the series converges as well.
I've proven that the sequence $\frac{n^{1/n}}{\ln(n)}$ continuous, positive, and decreasing over $(1, \infty)$. (I checked the derivative of it and the value is $\leq 0$).
Next, I used the limit comparison test, to compare $\frac{n^{1/n}}{\ln(n)}$ and $\frac{1}{\ln(n)}$, which diverges, meaning $\frac{n^{1/n}}{\ln(n)}$ also does.
Someone asked this question a while ago. But I remember somewhere in the lecture, there was once this theorem about if an element is removed/added from a convergent series (or something similar) the corresponding series would be convergent as well.
Does a series converge if its initial value is undefined?
Update: Can I actually use the Integral Test for this one?
Update 2: Sorry had a typo for $\frac{1}{\ln(0)}$
Thank You.
$\endgroup$ 54 Answers
$\begingroup$As noticed in the comments the series should be written as
$$\sum_{n=2}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$$
since for $n=1$ the expression is meaningless.
For the other issue, we have that for any convergent series since
$$\sum_{n_0}^\infty a_n=\sum_{n_0}^{n_1} a_n+\sum_{n_1+1}^\infty a_n$$
and the summation $\sum_{n_0}^{n_1} a_n$ is finite we have that
$$\sum_{n_0}^\infty a_n\quad \text{converges (or diverges)}\iff \sum_{n_1+1}^\infty a_n \quad \text{converges (or diverges)}$$
Your discussion for convergence is fine.
$\endgroup$ $\begingroup$Close to being an integral test
$$f(t)=t^{1/t}/\log t,\qquad f'(t)=O(1/t^2)+O(1/(t\log^2 t))$$$$f(2n)-f(2n+1)=\int_{2n}^{2n+1}f'(t)dt=O(1/(n\log^2 n))$$
$\endgroup$ $\begingroup$Thus $\sum_{n\ge 1} f(2n)-f(2n+1)$ converges absolutely.
To even be allowed to consider "$\sum_{n=0}^{+\infty} a_n$", the first thing is that all the $a_n$'s must be defined for $n \geq 0$. In your case, this is not satisfied, therefore the series has no sense.
If all is well-defined, we can say that ($\sum_{n=0}^{+\infty}$ converges) iff (for all $k \in \mathbb{N}$, $\sum_{n=k}^{+\infty}$ converges) : that means that the first terms are not important with regard to the convergence or the divergence of the series.
Finally, to your specific series (considering that the sum begins at $n=2$), you can see that $$\left| \frac{(-1)^n n^{1/n}}{\ln(n)}\right| \sim \frac{1}{\ln(n)}$$
so by comparison, the series is not absolutely convergent. But by the special criteria for alternating series, it is convergent. So the series is conditionally convergent.
$\endgroup$ $\begingroup$The series is an alternating series for $a_n=\frac{n^{1/n}}{\ln n}$. What you nee is to check if $a_n$ is decreasing and $\lim_{n\to\infty}a_n=0$.
(1). Since $n^{1/n}$ is decreasing when $n\ge 3$ and $\ln n$ is increasing, one has that $a_n$ is decreasing for $n\ge 3$.
(2). It is easy to check $\lim_{n\to\infty}a_n=0$.
Then by the Alternating Series Test, $\sum_{n=2}^\infty(-1)^na_n$ converges.
$\endgroup$More in general
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