Vtyjkyuk

How would you test for convergence with a series such as this using the alternating series test?

$1-frac12-frac13+frac14+frac15+frac16-frac17...-frac110+frac111...+frac115-....$

I have attempted to find a summation equation for this by grouping in different ways in order to use the alternating series test but I can't find a way to group these in a logical way. Is there a grouping that would allow for analysis of this series using the alternating series test?

I hope I'm interpreting this correctly: you have one $+$ term, then two $-$, three $+$, four $-$, etc. The start and end of a run of $2k+1$ $+$ terms are $1/(2 k^2 + k + 1)$ and $1/(2 k^2 + 3 k + 1)$ and the start and end of the next run of $2k+2$ $-1$ terms are $-1/(2 k^2 + 3 k + 2)$ and $-1/(2 k^2 + 5 k + 3$.
Thus the sum of these two runs is at least

$$ dfrac2k+12 k^2 + 3k + 1 - dfrac2k+22k^2+3k+2 = frac-k(2 k^2 + 3k + 1)(2k^2+3k+2)$$
and at most

$$ dfrac2k+12 k^2 +k + 1 - dfrac2k+22k^2+5k+3 = frac-k(2 k^2 + 3k + 1)(2k^2+3k+2) = frac6k+1(2k+3)(2k^2+k+1)$$

Thus in any case it is bounded in absolute value by $C/k^2$ for some $C$, and thus the series converges.

Define $$a_n = displaystylesum_k=fracn(n-1)2 + 1^fracn(n+1)2 frac 1k$$.

So for example, $a_1 = 1$, $a_2 = frac 12 + frac 13$, $a_3= frac 14 + frac 15 + frac 16$, etc.

It's easy to check that series you have written above is $sum_i=1^infty (-1)^i a_i$. It's easy to see that $a_i to 0$,since each fraction in the summation is less than $frac2n(n-1)+ 2$, and there are $n$ of them, so $a_i leq frac2nn(n-1) + 2$, which decreases to zero as $n to infty$.

You can also show that $a_n$ are decreasing as follows :
$a_n - a_n+1 = displaystylesum_k=fracn(n-1)2 + 1^fracn(n+1)2 frac 1k - displaystylesum_k=fracn(n+1)2 + 1^frac(n+2)(n+1)2 frac 1k$

What we do in this sum, is take terms whose difference is exactly $n$. More specifically, I claim that the above sum is the following:
$$
a_n - a_n+1 = frac-2(n+1)(n+2) + displaystylesum_k=fracn(n-1)2 + 1^fracn(n+1)2 left(frac 1k - frac1k+nright)
$$

As if to give an example, $a_3 - a_4 = frac 14 + frac 15 + frac 16 - frac 17 - frac 18 - frac 19 - frac 110 \ = frac 14 - frac 17 + frac 15 - frac 18 + frac 16 - frac 19 - frac 110$. Now you see why this sum holds.

Now, simplify the sums inside :
$$
a_n - a_n+1 = frac-2(n+1)(n+2) + displaystylesum_k=fracn(n-1)2 + 1^fracn(n+1)2 left(fracnk(k+n)right)
$$
Each term on the right hand side is greater than the term with $k = fracn(n+1)2$, and there are $n$ such terms, so:
$$
a_n - a_n+1 geq frac-2(n+1)(n+2) + nleft(fracnfracn(n+1)2left(fracn(n+1)2+nright)right)
$$
which after plenty of simplification becomes:
$$
a_n - a_n+1 geq frac-2(n+1)(n+2) + frac4(n+1)(n+3)
$$
which can be written as:
$$
a_n - a_n+1 geq 4 left(frac1(n+1)(n+3) - frac1(n+1)(2n+2)right)
$$
and the right hand side is positive when $n+3 leq 2n+2 $ i.e. when $n geq 1$.
For example, $a_3 - a_4 $ is predicted to be greater than $frac124$ by this estimate, and it's actual value is $frac3472520$, which is clearly greater than $frac 124$.

So, it follows that $a_n$ is a decreasing sequence which converges to $0$. Now, the alternate series test is applicable, I would think, and this shows that the given series converges conditionally.

Please point out any mistake if found.
$()$