Is there an error in the following statement of the Integral Test (from David Brannan's Mathematical Analysis)?
The author requires the integrand (equivalently, the general term of the series) to converge to zero for the test to be applied. Is this correct? For if the integrand converges to zero, then the series converges, so there is no need to apply the Integral Test, is there?
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This may help: Note that the convergence test states that IF a series converges, THEN the limit of $f(n)$/"the integrand" is $zero$.
The converse does not necessarily follow.
That is, if the $\lim_{n\to \infty} f(n) = 0,\;$ we cannot conclude that the series converges. We don't know. Recall the example of $f(n) = \large\frac 1n.\;$ It's limit as $n\to \infty$ is $0$, but the series $\sum_{n^\infty} f(n)$ does not converge.
It's cases like this when the integral test comes in handy.
Indeed, the condition that $\lim_{n \to \infty} f(n) = 0$ must be met for the integral test to apply.
ADD The condition that $a_n\not \to 0$ is sufficient for a sequence $a_n$ to fail to have a sum $$\sum_{n=0}^\infty a_n$$ That is, if $a_n\not\to 0$ then $\sum a_n$ fails to exists.
This implies, by contrapositiveness, that the condition $a_n\to 0$ is necessary for the sequence to have a sum, but it is not sufficient. That is, if $\sum a_n$ exists, then $a_n\to 0$.
However, there is a canonical example where we see how this fails:
$$\frac 1 n\to 0\text{ however } \sum \frac 1 n $$ fails to exist.
Observe that because $f$ is positive and decreasing
$$\sum_{k=2}^n f(k)\leq \int_1^n f \leq \sum_{k=1}^{n-1}f(k)$$
From the first side of the above, and assuming $$\int_1^n f $$ is bounded above by some number $R$, we get that $$\sum_{k=2}^n f(k)\leq R$$ for each $n$.
Since $f$ is positive, the series is monotone increasing, so that by the monotone convergence theorem, the result follows.
On the other hand, the rightmost inequality implies that if $$ \int_1^n f $$ diverges, so does $$\sum_{k=1}^{n-1}f(k)$$
by direct comparison.