SOLUTION: Determine if the limit exists as n goes to infinity: {{{ 1/(n+1)+1/(n+2)}}} +...+ {{{1/(2n-1) + 1/(2n)}}}. If the limit exists then find its value.

Algebra ->  Sequences-and-series -> SOLUTION: Determine if the limit exists as n goes to infinity: {{{ 1/(n+1)+1/(n+2)}}} +...+ {{{1/(2n-1) + 1/(2n)}}}. If the limit exists then find its value.      Log On


   

Question 1103369: Determine if the limit exists as n goes to infinity:
+1%2F%28n%2B1%29%2B1%2F%28n%2B2%29 +...+ 1%2F%282n-1%29+%2B+1%2F%282n%29.
If the limit exists then find its value.
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
Let +S%5Bn%5D+=+1%2F%28n%2B1%29+%2B+1%2F%28n%2B2%29+ + … + +1%2F%282n-1%29+%2B+1%2F2n+
Consider +nS%5Bn%5D+=+n%2F%28n%2B1%29+%2B+n%2F%28n%2B2%29+ + … + +n%2F%282n-1%29+%2B+n%2F2n+
For +n%3E=1+:
+n%2F%28n%2B1%29+%3C+1+
+n%2F%28n%2B2%29+%3C+1+
: :
+n%2F%282n%29+%3C+1+
That's n terms, each of which is less than 1.
+nS%5Bn%5D+%3C+n+
which implies
+S%5Bn%5D+%3C+1+
For the lower bound: +nS%5Bn%5D+%3E=+n%2F2+ (all but the last term are > n/2)
+S%5Bn%5D+%3E+1%2F2+
So +1%2F2+%3C+S%5Bn%5D+%3C+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%5Bn%5D+-+S%5Bn-1%5D+=+1%2F%282n%2A%282n-1%29%29++ which goes to 0 monotonically as +n —> +infinity+. This means +S%5Bn%5D+ increases by smaller amounts as n increases, and the series converges.