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I looked this up on Wolfram Alpha and it said it converged to $100.578$. Is this correct? Much more importantly, how would I solve the question please? I am at pre-uni level (this is an old interview question) so I'm afraid even basic techniques from first-year analysis aren't allowed, I think.

Thank you

HK Lee
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John Smith
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  • Yes, the sum converges to that value, and it is known that it converges for any $t>1$ (that is, $t=1.01$ in this case). The value is given by the zeta function, which is famous, but somebody else here may tell you more, because I know almost nothing about it, not even how to calculate it. – Sarvesh Ravichandran Iyer Nov 19 '16 at 07:52
  • You claim this is an interview question; where did you get the interview question? I would say if you aren't allowed to use things you know (e.g., either of the answers below), then the rules need to be spelled out more clearly. – Clayton Nov 19 '16 at 08:12
  • A website claiming to have a list of past oxbridge interview questions - it seemed interesting so I tried it – John Smith Nov 19 '16 at 08:42

2 Answers2

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It depends on where in the world you are as to what pre-university allows.

In the UK you might reasonably argue that $$\displaystyle \int_1^\infty \dfrac{1}{x^{1.01}}dx \lt \sum_1^\infty \dfrac{1}{n^{1.01}} \lt 1+\int_1^\infty \dfrac{1}{x^{1.01}}dx$$ with a sketch to justify it.

enter image description here

The points you could make in the interview include:

  • The left hand side is then $100$ and the right $101$, setting bounds on the sum.

  • Careful examination of the sketch might take you to the sum being close to $100.5$

  • A convexity argument would take this to slightly more than $100.5$.

  • Some of the "slightly more" comes in the early terms which you might be able to calculate explicitly.

Henry
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This is standard consequence of p-series test: The series $\sum_{n=1}^\infty \frac{1}{n^p}$ converges for $p>1$ and diverges for $p\leq1$. The convergence can be proved by Cauchy Condensation Test and by using convergence of geometric series.

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vidyarthi
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