Which convergence test do I use?

If \sum a_n converged to a sum S, its partial sums S_n would converge to S, and so would S_{n-1} (the same sequence, shifted one index — a shift by one term can't change what a sequence converges to). But a_n = S_n - S_{n-1} for n \geq 2, so by the sequence limit laws, a_n \to S - S = 0. Contrapositive: a_n \not\to 0 forces \sum a_n to diverge.

For p\le 0 the terms 1/n^p don't even shrink to zero (they're constant or growing), so the series diverges by the divergence test alone. For p>0, f(x)=1/x^p is continuous, positive, and decreasing on [1,\infty), so the Integral Test applies: \int_1^{\infty} x^{-p}\,dx converges exactly when p>1 (a direct antiderivative computation — for p=1 the antiderivative is \ln x, unbounded as x\to\infty; for p\ne 1 it's \frac{x^{1-p}}{1-p}, which has a finite limit at infinity only when the exponent 1-p is negative, i.e. p>1). The series and the integral share that same threshold.

Since every term is nonnegative, both series' partial sums are nondecreasing sequences, so each either converges or grows without bound — there's no third option. If \sum b_n converges to L, every partial sum of \sum a_n satisfies \sum_{n=1}^{N} a_n \le \sum_{n=1}^{N} b_n \le L, so \sum a_n's partial sums are nondecreasing and bounded above, forcing convergence. The second bullet is the contrapositive of the first.

Since the ratio a_n/b_n\to L>0, eventually \frac{L}{2} < \frac{a_n}{b_n} < \frac{3L}{2}, which rearranges to \frac{L}{2}\,b_n < a_n < \frac{3L}{2}\,b_n for all large enough n. If \sum b_n converges, the right inequality and the Direct Comparison Test give \sum a_n converges too (comparing against \frac{3L}{2}b_n, itself a convergent series since it's a constant multiple of one). If \sum b_n diverges, the left inequality and Direct Comparison give \sum a_n diverges too (comparing \frac{L}{2}b_n \le a_n, and a constant multiple of a divergent nonnegative series still diverges). Swapping the roles of a_n and b_n (using 1/L as the limiting ratio the other way) gives the same conclusion starting from either series, so the two series stand or fall together.

Suppose L<1. Pick r with L<r<1. Since the ratio's limit is L, eventually (say for n\ge N) |a_{n+1}/a_n| < r, so |a_{n+1}| < r|a_n|. Applying this repeatedly from n=N gives |a_{N+k}| < r^k|a_N| for every k\ge 0 — the tail is dominated term by term by a convergent geometric series (ratio r<1) scaled by |a_N|. The Direct Comparison Test then gives \sum|a_n| converges, i.e. absolute convergence. If L>1, the same argument with r between 1 and L shows |a_{n+1}|>|a_n| eventually, so the terms are eventually increasing in magnitude and cannot shrink to 0 — the Divergence Test then rules out convergence. L=1 is inconclusive because both a convergent p-series (p=2) and the divergent harmonic series (p=1) produce a ratio limit of exactly 1, so the test alone cannot distinguish them.

Suppose L<1. Pick r with L<r<1. Eventually \sqrt[n]{|a_n|}<r, so |a_n|<r^n for all large n — the tail is dominated term by term by a convergent geometric series with ratio r<1. The Direct Comparison Test then gives \sum|a_n| converges. If L>1, eventually \sqrt[n]{|a_n|}>1, so |a_n|>1 infinitely often (in fact eventually always), meaning the terms cannot shrink to 0 — the Divergence Test then rules out convergence. L=1 is inconclusive for the same reason as the Ratio Test: both a convergent p-series and the divergent harmonic series give a root limit of exactly 1.

Because f is decreasing, on each unit interval [k,k+1] the function never exceeds its left value: f(x) \le f(k) = a_k for x in [k,k+1], so \int_k^{k+1} f(x)\,dx \le a_k. Summing from k=N to k=M-1 gives \int_N^{M} f(x)\,dx \le \sum_{n=N}^{M-1} a_n. The same decreasing property also gives f(x) \ge f(k+1) = a_{k+1} on [k,k+1], so a_{k+1} \le \int_k^{k+1} f(x)\,dx, and summing that from k=N to M-1 gives \sum_{n=N+1}^{M} a_n \le \int_N^{M} f(x)\,dx. Together, the partial sums and the running integral are trapped against each other closely enough that one grows without bound exactly when the other does — since f\ge 0 makes both \int_N^t f(x)\,dx and the partial sums nondecreasing in the first place, "trapped and nondecreasing" forces them to converge or diverge together as M\to\infty.

R_n = \sum_{k=n+1}^{\infty} a_k is itself a tail series with the same integral-test bounding argument applied starting at N=n+1: the same inequalities used to prove the Integral Test give \int_{n+1}^{\infty} f(x)\,dx \le \sum_{k=n+1}^{\infty} a_k (the lower integral bound on each unit strip) and \sum_{k=n+1}^{\infty} a_k \le \int_{n}^{\infty} f(x)\,dx (the upper bound, re-indexed one step earlier since a_{n+1}\le\int_n^{n+1}f). Both tail integrals converge whenever the original one does, since a convergent improper integral's tail beyond any finite point is itself a convergent improper integral.

Consider the even partial sums S_{2n} of \sum(-1)^{n+1}b_n. Grouping consecutive pairs, S_{2n} = (b_1-b_2)+(b_3-b_4)+\cdots+(b_{2n-1}-b_{2n}) — every grouped pair is nonnegative since b_n is decreasing, so S_{2n} is nondecreasing in n. Regrouping the same sum as S_{2n}=b_1-(b_2-b_3)-(b_4-b_5)-\cdots-b_{2n} shows S_{2n}\le b_1 for every n, since each parenthesized pair is again nonnegative and is being subtracted. A nondecreasing sequence bounded above converges, so S_{2n}\to S for some S. The odd partial sums satisfy S_{2n+1}=S_{2n}+b_{2n+1}, and b_{2n+1}\to 0, so S_{2n+1}\to S as well — both subsequences converge to the same limit, which forces the whole sequence of partial sums to converge to S. The other sign pattern is the same series multiplied by -1, which converges whenever the original one does.

From the proof of the Alternating Series Test, S lies between any two consecutive partial sums: S_n and S_{n+1}=S_n+(-1)^{n+2}b_{n+1} straddle S, since the even partial sums increase to S from below while the odd ones decrease to S from above (or vice versa, depending on parity). Being trapped between S_n and S_{n+1} means |S-S_n| is at most the gap between those two consecutive partial sums, which is exactly |S_{n+1}-S_n| = b_{n+1}.