Vtyjkyuk

Show that the series converges, but not absolutely. $sum_n=1^infty( $exp$(frac(-1)^nn)-1)$.

My Try:

Let $a_n=$exp$(frac(-1)^nn)-1$. I was going to use alternating series test because the sequence $a_n$ is alternating. But $a_n$ is not decreasing to $0$. So, I am stuck now. Can anybody please give me a hint?

You may just use the fact that, as $x to 0$, using the Taylor expansion,
$$
e^x=1+x+O(x^2)
$$ giving, for some $n_0geq1$,
$$
sum_ngeq n_0left(exp left(frac(-1)^nnright)-1right)=sum_ngeq n_0frac(-1)^nn+sum_ngeq n_0Oleft(frac1n^2right).
$$ The latter series is absolutely convergent and the series $displaystyle sum_ngeq n_0frac(-1)^nn$ is conditionally convergent. It gives the desired result.

PART 1: ESTABLISHING CONVERGENCE

From the Mean Value Theorem we have

$$e^(-1)^n/n-1=frac(-1)^nn+e^xi_nfrac1n^2$$

for $0<|xi_n|<frac1n$. Then, we have

$$beginalign
sum_n=1^inftyleft(e^(-1)^n/n-1right)&=sum_n=1^inftyleft(frac(-1)^nn+e^xi_nfrac1n^2right) tag 1
endalign$$

we recall that the alternating harmonic series $sum_n=1^inftyfrac(-1)^nn=log 2$ converges. And we note that since $|e^xi_n|<e$ for $0<|xi_n|<frac1n$, then the series

$$beginalign
sum_n=1^inftye^xi_nfrac1n^2 &le esum_n=1^inftyfrac1n^2\\
&=fracpi^2,e6
endalign$$

also converges. Finally, since the sum of two convergent series is a convergent series, the series on the left-hand side of $(1)$ converges.

PART 2: SHOWING CONVERGENCE IS CONDITIONAL

Now, we observe that the series of absolute values is bounded below as

$$beginalign
sum_n=1^inftyleft|e^(-1)^n/n-1right|&=sum_n=1^inftyleft|frac(-1)^nn+e^xi_nfrac1n^2right|\\
&ge frac12,sum_n=1^inftyfrac1n
endalign$$

which diverges since the harmonic series diverges. Thus, the series of interest is not absolutely convergent, only conditionally convergent.