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    Hi, Can someone help me with q8 of Jan 2009

    Paper: http://pmt.physicsandmathstutor.com/...0FP2%20OCR.pdf

    MS: http://pmt.physicsandmathstutor.com/...0FP2%20OCR.pdf

    So I get that the first 2 questions were asking about upper and lower bounds. I just don't get how to get the answer in part 3.

    Part 4 doesn't seem to make sense I'm pretty sure that sum to infinity of 1/r is a convergent series.

    Thanks
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    Part 3 follows upon comparing the two inequalities you've already found! Try adding +1 to both sides of the first inequality and 1/(n+1) to both sides of the second inequality. Then, using this result (wedging the sum between two closed-form functions), you can deduce that the series DOES NOT converge, because as n tends to infinity, the leftmost and rightmost sides of the triple inequality tend to infinity.

    In fact you only need to look at the left inequality, stating that the series is greater than ln(n+1) + 1/(n+1) for a given n. As n tends to infinity the function ln(n+1) + 1/(n+1) gets larger and larger without bound (i.e. it tends to infinity). Therefore, since the series is always greater than this function, it too must get larger and larger without bound - in other word, it diverges.

    This therefore proves a rather surprising result, that the sum of the reciprocals of the integers (1 + 1/2 + 1/3 + 1/4 + ...) never converges to a finite value, instead shooting off to infinity, even though the amounts you add as n gets large become really tiny!
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    (Original post by ChemBoy1)
    Hi, Can someone help me with q8 of Jan 2009

    Paper: http://pmt.physicsandmathstutor.com/...0FP2%20OCR.pdf

    MS: http://pmt.physicsandmathstutor.com/...0FP2%20OCR.pdf

    So I get that the first 2 questions were asking about upper and lower bounds. I just don't get how to get the answer in part 3.

    Part 4 doesn't seem to make sense I'm pretty sure that sum to infinity of 1/r is a convergent series.

    Thanks
    \sum \frac{1}{r} is a famous divergent series, but come back to that later.

    Anyhow, note that from (i) we can add 1 to both sides and get 1+\frac{1}{2}+\frac{1}{3}+...+ \frac{1}{n+1} < \ln(n+1)+1 and the LHS is precisely \displaystyle \sum_{n=1}^{n+1} \frac{1}{r} hence we obtain an upper bound on it; \displaystyle \sum_{n=1}^{n+1} \frac{1}{r} < \ln (n+1) +1

    Now use part (ii) to do something similar and obtain a lower bound for it. Put the two together.

    As for the last part, observe what happens to the lower bound when n \rightarrow \infty, and link it back to the sum.
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    (Original post by chailatte2407)
    ...
    Would be best if you don't outright answer the questions and let OP have a go by themselves again after a few hints.
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    Thanks RDK Games, I got a better understanding now, once I added 1 and 1/n+1 to RHS and LHS respectively it made sense. Although, the converging series question still puzzles me. I keep looking at the graph and it shows that it coverges, could you help me out on this? How can a converging looking graph actually be diverging?
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    (Original post by ChemBoy1)
    Thanks RDK Games, I got a better understanding now, once I added 1 and 1/n+1 to RHS and LHS respectively it made sense. Although, the converging series question still puzzles me. I keep looking at the graph and it shows that it coverges, could you help me out on this? How can a converging looking graph actually be diverging?
    The value of the graph \frac{1}{x+1} tends to 0, but the actual area between it and the x-axis does not converge. This is the distinction - you're proving something to do with area.
 
 
 
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