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I never fully got the Integral Test when we had a lecture on it a few weeks ago, so I would appreciate an explanation for why we have (provided $f$ is a non-negative decreasing function):

$$\int^k_1 f(x) dx \le \sum^k_{n=1}f(n) \le \int^k_0f(x)dx$$

When as far as I see it, the proof is reliant on the fact that:

$$f(k) \le \int^k_{k-1}f(x)dx \le f(k-1)$$

As would be clear from a diagram, but this leads me to conclude that if we let

$$\int^k_1f(x)dx=\int^2_1f(x)dx+ \int^3_2f(x)dx + ... + \int^k_{k-1}f(x)dx \le f(1) + f(2)+...+f(k-1)$$

So we should add $f(x)$ or $f(k)$ I suppose to both sides, but this suggests that the top inequality is incorrect? I can only imagine I must have copied it down wrong or I have made some error in trying to prove it.

Then to clarify, the other side of the inequality is from a similar idea, that:

$$\int^k_0f(x)dx=\int^1_0f(x)dx + \int^2_1f(x)dx+ \int^3_2f(x)dx + ... + \int^k_{k-1}f(x)dx \ge f(1) + f(2)+...+f(k-1) + f(k)$$

Which does agree with what I have at the top.

Then I guess my second "question" is that is then enough to say that if the limit of the integral $f(x)$ as $k \to \infty$ is below infinity, then the sequence of partial sums is bounded by the RHS of the top and so must converge by the Axiom of Completeness? Then I am not sure what to conclude in regard to the left hand side, if the integral on the left is infinity then it diverges?

Thanks!

One last thing, can the RHS be written as:

$$\int^k_1 f(x)dx + f(1)$$

And is there any benefit to doing so?

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On your first inequality with the $k$ integrals, add $f(k)$ to both sides. Since $f(k)$ is non-negative,

$$ \int_{1}^k f(x)dx \leq \int_{1}^k f(x)dx + f(k) \leq f(1)+f(2)+\dots+f(k-1)+f(k), $$ so that $$ \int_{1}^k f(x)dx \leq \sum_{n=1}^k f(n). $$

Regarding your second question, take the limit as $n$ goes to infinity on both sides of your topmost inequality. By assumption both of those integrals converge, so that the infinite sum in the middle converges by the squeeze theorem. And if either integral diverges, then the sum in the middle also diverges (also by squeeze theorem).

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    Ah ok, so it is already smaller than $\sum^k_{n=1}f(n)$ but there would also be no harm in adding $f(k)$ to the LHS because it will still be $\le$? – 2017-01-05
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    We added $f(k)$ to both sides so that the sum on the RHS would go up to $f(k)$. Since $f(k)$ is non-negative it doesn't change the inequality, and so that the integral up to k is $\leq$ the integral up to k, plus $f(k)$. – 2017-01-05