It can't be that 
makes the series converge, because this would introduce a zero in the denominator when 
. For a similar reason, 
would involve an indeterminate term of 
.
That leaves checking what happens when
. First, consider the function
and its derivative
has critical points at 
and 
. (These never coincide because we're assuming , so it's always the case that 
.) Between these two points, say at 
, you have 
, which is positive regardless of the value of 
. This means is increasing on the interval 
.
Meanwhile, if
- and let's take 
as an example - we have 
, which is negative for all . This means is decreasing for all , which shows that 
is a decreasing sequence for all 
, where 
is any sufficiently large number that depends on .
Now, it's also the case that for (and in fact all 
),
So you have a series of a sequence that in absolute value is decreasing and converging to 0. The alternating series then says that the series must converge for all.
For the second question, recall that
(note that the above is true for even
only - it wouldn't be too difficult to change things around if is odd)
It follows that
Subtracting
from 
, you have
as required. Notice that assuming is odd doesn't change the result; the last term in ends up canceling with the corresponding term in regardless of the parity of .
That leaves checking what happens when
and its derivative
Meanwhile, if
Now, it's also the case that for
So you have a series of a sequence that in absolute value is decreasing and converging to 0. The alternating series then says that the series must converge for all
For the second question, recall that
(note that the above is true for even
It follows that
Subtracting
as required. Notice that assuming