I couldn't find an easy to way to check for the convergence of the series. The sum is supposed to be from 2 to infinity and there is a little mistake in the pic.
Since $\frac1{n-1}$ diverges by comparison test with $\frac1n$. Would it be right to say that the whole series is divergent because one of the part of the sum is divergent ? Do l still need to prove that the other part diverges or not ?
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$\begingroup$Manipulating infinite sums directly and splitting them up like that is hairy business, and showing when it can be done and why is actually quite theoretically heavy. But fortunately, we don't really need that here.
Clearly, for any finite natural number $k\geq 2$, we have$$ \sum_{n = 2}^k\frac1{n-1} \leq \sum_{n = 2}^k\frac{1}{n-\sqrt n} $$You can use the exact same algebraic manipulation as you have done to show this, but it's even easier than that: The terms on the right have smaller numerators, so they must be larger.
It is not difficult to use our knowledge that the left side goes to $\infty$ as $k$ grows to prove that the right side goes to $\infty$ as $k$ grows. One could go all out and use an $\varepsilon$-$N$ style proof of this, and if that's something that sounds familiar to you, then I think you should do just that as an exercise. It may actually be even easier to do it in general: Take two sequences $a_k, b_k$ with $a_k\to \infty$ and $a_k\leq b_k$, then prove that $b_k\to \infty$.
If it doesn't sound familiar to you, then this is rather difficult to prove formally (as the formal notions of limit at infinity and divergence to infinity depend on $\varepsilon$-$N$ in the first place). But as a hand-wavey proof, pointing out that the left-hand side grows monotonically without bound as $k$ grows should be enough to convince most that the right-hand side grows monotonically without bound too. I mean, how could it not?
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