mjhjmtu

Prove that given sequence $$langle f_nrangle =1-frac12+frac13-frac14+.....+frac(-1)^n-1n$$

is a Cauchy sequence

My attempt :
$|f_n-f_m|=Biggl|dfrac(-1)^mm+1+dfrac(-1)^m+1m+2cdotsdots+dfrac(-1)^n-1nBiggr|$

using $ m+1>m implies dfrac1m+1<dfrac1m $

$|f_n-f_m|le dfrac1m+dfrac1m+dfrac1mcdotscdotsdfrac1m$

$|f_n-f_m|ledfracn-mm$

I don't know if I am proceeding correctly or if I am, how to proceed further, any hint would be really helpful .

If you ignore the signs of the terms,
the result diverges.
So you can't do that.

$f_n
=sum_k=1^n dfrac(-1)^kk
$
so,
if $n > m$,
$f_n-f_m
=sum_k=m+1^n dfrac(-1)^kk
=sum_k=1^n-m dfrac(-1)^k+mk+m
=(-1)^msum_k=1^n-m dfrac(-1)^kk+m
$.

If
$n-m$ is even,
so $n-m = 2j$,
then

$beginarray\
f_n-f_m
&=(-1)^msum_k=1^2j dfrac(-1)^kk+m\
&=(-1)^msum_k=1^j left(dfrac(-1)^2k-12k-1+m+dfrac(-1)^2k2k+mright)\
&=(-1)^msum_k=1^j (-1)^2k-1left(dfrac-12k-1+m+dfrac12k+mright)\
&=(-1)^msum_k=1^j (-1)^2k-1left(dfrac(2k-1+m)-(2k+m)(2k-1+m)(2k+m)right)\
&=(-1)^m+1sum_k=1^j left(dfrac-1(2k-1+m)(2k+m)right)\
&=(-1)^msum_k=1^j left(dfrac1(2k-1+m)(2k+m)right)\
textso\
|f_n-f_m|
&=sum_k=1^j left(dfrac1(2k-1+m)(2k+m)right)\
&=sum_k=1^jdfrac14 left(dfrac1(k-frac12+fracm2)(k+fracm2)right)\
&lt dfrac14sum_k=1^j left(dfrac1(k-1+fracm2)(k+fracm2)right)
quadtextthis is the sneaky part\
&lt dfrac14sum_k=1^j left(dfrac1k-1+fracm2-dfrac1k+fracm2right)\
&= dfrac14 left(dfrac1fracm2-dfrac1j+fracm2right)\
&= dfrac12 left(dfrac1m-dfrac12j+mright)\
&= dfrac12 left(dfrac1m-dfrac1nright)\
&< dfrac12m\
&to 0 text as m to infty\
endarray
$

If $n-m$ is odd,
the sum changes
by at most $frac1n$
so it still goes to zero.

Hint :

$$frac12n-1-frac12n=frac1(2n-1)2nleq frac1(n-1)n= frac1n-1-frac1n $$