values for $x$ in which the series converges I have seen some videos and looked up different examples on the internet but non seem to help me start an approach to my problem which is the following : find all the values $a$ in which the series converges:
$$ 
  \sum_{n=1}^\infty \left(\frac a{n+2}-\frac1{n+4}\right)
$$ 
most of the examples ive seen have to deal with powers, which i understood after seeing alot of examples, but havent come across an example that helps me with my problem. Thanks for any help!
 A: Taken separately, the two terms form a diverging harmonic series, and except for a few terms you get $(a-1)H_n$. The only way to counteract that is to make $a=1$ so that cancellation enters into play and only the "few terms" remain.
Now,
$$\frac1{n+2}-\frac1{n+4}=\frac2{(n+2)(n+4)},$$ leading to a converging series.
A: your finit sum is given by the following formula $$- \left( n+3 \right) ^{-1}- \left( n+4 \right) ^{-1}+ \left( \alpha-1
 \right) \Psi \left( n+3 \right) +{\frac{7}{12}}- \left( \alpha-1
 \right)  \left( 3/2-\gamma \right) 
$$
A: We have
$$u_n=\frac{a}{n}\frac{1}{1+\frac{4}{n}}-\frac{1}{n}\frac{1}{1+\frac{2}{n}}$$
$$=\frac{a}{n}(1-\frac{4}{n}+....$$
$$-\frac{1}{n}(1-\frac{2}{n}+....$$
it converges if the coefficient of $\frac{1}{n}$ in Asymptotic expansion is zero, which gives
$$a-1=0.$$
A: Here is the basic idea:  For any value of $a$, we have
$${a\over n+2}-{1\over n+4}={(a-1)n+(4a-2)\over n^2+6n+8}$$
If $a\not=1$, then
$${(a-1)n+(4a-2)\over n^2+6n+8}\approx{a-1\over n}$$
for large $n$, which means the infinite sum is comparable to the (divergent) harmonic series when $a\not=1$.  Can you see what happens in the other case, when $a=1$?
There are various ways to make the comparison rigorous.  If you need additional help, please ping with a comment.
A: Notice that
$$\frac a{n+2}-\frac1{n+4}=\frac{(a-1)n+4a-2}{(n+2)(n+4)}$$
Now try limit comparison to $\frac1n$.
