Question 1103369
Let {{{ S[n] = 1/(n+1) + 1/(n+2) }}} + … + {{{ 1/(2n-1) + 1/2n }}}

Consider {{{ nS[n] = n/(n+1) + n/(n+2) }}} + … + {{{ n/(2n-1) + n/2n }}}
For {{{ n>=1 }}}:
{{{ n/(n+1) < 1 }}}
{{{ n/(n+2) < 1 }}}
:    :
{{{ n/(2n) < 1 }}}

That's n terms, each of which is less than 1. 
 {{{ nS[n] < n }}} 
which implies
 {{{ S[n] < 1 }}}   

For the lower bound: {{{ nS[n] >= n/2 }}}  (all but the last term are > n/2)
 {{{ S[n] > 1/2 }}}

So   {{{ 1/2 < S[n] < 1 }}}  and the series has a limit as  {{{ n }}} —> {{{ infinity }}} .  
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To find the limit value, I cheated and used a one-line Perl script:
perl -e '$m=10000000; for($i=$m, $s=0; $i<(2*$m); $i++) { $s += 1/($i+1); } print "$s\n";'
0.693147155559907
perl -e '$m=100000000; for($i=$m, $s=0; $i<(2*$m); $i++) { $s += 1/($i+1); } print "$s\n";'
0.693147178059741

Which looks like ln(2).      

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EDIT:  I also should have shown that the series terms get smaller as {{{ n }}} —> {{{ infinity }}}
{{{ S[n] - S[n-1] = 1/(2n*(2n-1))  }}} which goes to 0 monotonically as {{{ n}}} —> {{{ infinity }}}.  This means {{{ S[n] }}} increases by smaller amounts  as n increases, and the series converges.