341look up Harmonic Series and its divergence
Perhaps you are looking for:
\lim_{n \rightarrow \infty} \frac{1}{1+n} + \frac{1}{2+n} + ... + \frac{1}{2n}
sum the series
1+ 1/2 + 1/3 + 1/4..........
i used integration ....but that doesn't help as log {0} doesn't exist .....please help.....please give an explanation with your answer
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11 Answers
341
1sorry sir i didn't get harmonic series and its divergence ....
what is it ?
1The given series is 1,1/2,1/3,.....,1/n
Tn=1/n
Sn =∫Tn =∫1/n =log(n)
n>0
24we are looking for
S= x/1 +x2/2 +x3/3 +.............+xn/n
diff. => S' =x+x2+x3+......+xn=(1-xn)/(1-x)
1
=>S=∫1-xn)/(1-x) dx
0
1if u put x=2 in the 'S' u hv given the ques changes.....
S= x/1 +x2/2 +x3/3 +.............+xn/n
1@ honey as n →∞ ...we have log (∞)......we may say =∞
but 1 + 1/2 + 1/3 + ......= ∞ .......isn't something wrong .....all nos. are less than or equal to 1 ......and that too there no. extend upto infinity ,...ie upto zero ......i find something strange
there sum must be finite
341@rohan: I will try one more time.
look up wikipedia for harmonic series.
this applies to the other guys who have posted in this thread too
66This series is the so called harmonic series. This series diverges (which means that sum to n terms increases as n increases) and hence we cannot sum it.
1http://demonstrations.wolfram.com/TheSumOfTheHarmonicProgressionAsAnIntegral/