Find a closed form for the nth partial sum of the series, and determine whether the series converges...?

Question

Hey everybody, here's a question from my Calc2 book:

--Find a closed form for the nth partial sum of the series, and determine whether the series converges. If so, find its sum:

ln(1/2) + ln(2/3) + ln(3/4) + ... + ln(k/(k+1)) + ...

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Any ideas? I haven't had problems deriving the closed form before, but I've never encountered an open formula like this one, so I don't really know where to start. Makes me feel like I don't understand the concept at all! Any help with this would be greatly appreciated. Thanks! 8^0

Answer 1

Remember that the sum of logs is the log of the product:

ln(a) + ln(b) = ln(a*b)

So: ln(1/2) + ln(2/3) + ln(3/4)..... ln(k/(k+1)) =
ln((1/2) * (2/3) * (3/4) * ... * k/(k+1))

You can see that all the previous denominators are cancelled by the next numerator so:

ln(1/2) + ln(2/3) + ln(3/4)..... ln(k/(k+1)) = ln(1/(k+1)) = -ln(k+1)

From this, the series does not converge, but it does diverge slowly.

Answer 2

Use the rules of logarithms: a sum of logs is the log of the product, so:

ln(1/2) + ln(2/3) + ln(3/4) + ... + ln(k/(k+1)) = ln[(1/2) * (2/3) * (3/4) *...* (k/(k + 1))] = ln(1/(k+1)). Thus the answer is, the kth partial sum is ln(1/(k+1)).

Answer 3

Use your log properties to simplify!

lnA + lnB = ln (AB)

you will see a NICE canceling.. 1/2 * 2/3 * 3/4 * 4/5.....

Look at partial sums:
s2 = ln (1/3)
s3 = ln (1/4)
s4 = ln (1/5)
s5 = ln (1/6)

... unfortunately the sums will DIVERGE, because as n goes to infinity, 1/n goes to zero. AND finally ln(approaching 0) goes to negative infinity.

Answer 4

For this sequence concern, lets favor to continually start up off with the divergence attempt: this states that if the shrink of a chain isn't equivalent to 0, then the sequence diverges (in diverse words, if it has a shrink diverse than 0, it diverges). If the shrink is resembling 0, we gained't tell, and could use yet another attempt (evaluation, crucial, ratio, etc). yet you are able to favor to continually do the divergence attempt first. (For calc II, the divergence attempt is largely one issue it extremely is given to you; a truth of existence. you do now no longer could favor to be waiting to coach it). So, The shrink as n ideas infinity of (3n-a million)/(5n+7). properly, in case you recognize your math guidelines, then you definately definately will understand that once the polynomial on the precise has the similar degree because of the very truth the polynomial on the bottom, the shrink is largely the ratio of the coefficients (therefore, the shrink is 3/5). you are able to also practice this by technique of creating use of L'medical institution's rule: take the by technique of-man made from the precise (3) and the by technique of-man made from the bottom (5) and hit upon the shrink of that, it extremely is likewise the shrink of the unique (yet pay interest, you are able to in elementary words attempt this even as the shrink of the unique is infinity/infinity or 0/0). in any case, the shrink is 3/5. 3/5 is of route now no longer equivalent to 0, so this sequence diverges. and because it diverges, there is not any sum! severe-nice :D choose this enables!