SOLUTION: (1/log2)-(1/log 3)+(1/log4)-..... what will be the nth term? whether it is right to write the series from n=2 to infinity (1/log n)

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Question 813321: (1/log2)-(1/log 3)+(1/log4)-..... what will be the nth term?
whether it is right to write the series from n=2 to infinity (1/log n)
Answer by Edwin McCravy(20066) About Me  (Show Source):
You can put this solution on YOUR website!
(1/log2)-(1/log 3)+(1/log4)-..... what will be the nth term?
The nth term is +%28-1%29%5En%2Flog%28%28n%29%29+ for n = 2,3,4,...

or 
 
the nth term will be +%28-1%29%5E%28n%2B1%29+%2Flog%28%28n%2B1%29%29 for n = 1,2,3,...

or 

the nth term will be ++%28-1%29%5E%28%28n%29%29%2Flog%28%28n%2B2%29%29+ for n = 0,1,2,...

It depends on what value you want to start the dummy variable n.

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whether it is right to write the series from n=2 to infinity (1/log n)

It is OK if you include the sign-alternating factor (-1)n, but not
if you don't include it.

S = sum%28+%28%28-1%29%5En%2Flog%28%28n%29%29%29%2Cn=2%2Cinfinity%29 is OK

An alternating series converges (at least conditionally) if the
nth term approaches 0.

Edwin