Erm. Why the downvote, if I may? (Is there something wrong with my answer, and if so what -- so that I can fix it?)

(If it is the presence of the beige/gray boxes... these are "spoilers", clicking or hovering over them will reveal the text. If it is something else, please, comment on it.)

What is wrong with my answer?

@DonLarynx It's too lose an upper bound to conclude: you are only showing $(a_n)_n$ is bounded in your first points, which is exponentially weaker than the tight geometric bound. I actually do not understand what you are arguing in your points (b) and (c). Also, tangentially I don't really what this has to do with the answer I gave, i.e. why you are commenting here (?)..

Connecting the dots of the sequence, the graph looks like a concave down curve and therefore converges due to the monotone convergence theorem (bounded below by zero). Agree?

Because I need help, I'm a student.

What does $(a_n)_n$ mean? Did you mean $a_n$? $(a_1)_1$, $(a_2)_2$, $(a_n)_n$ does not make sense.

But the same "concavity argument" would apply to $a_{n+1}=a_n=c >0$, and surely the corresponding series does not converge. As for your last question: $(a_n)_{n\geq 0}$ is a sequence, $a_n$ is the $n$-th term of the sequence.

Recall $a_{n+1}$ will never be equal to $a_n$ because the ratio $\frac{a_{n+1}}{a_n}$ is less than 1.

I know. This is what my proof is based on, but your argument is actually too weak: from the original argument, you prove something too weak to conclude (you only basically use $\lvert a_{n+1}\rvert < \lvert a_{n}\rvert$. This is not enough, e.g. consider $a_{n+1} = (1-\frac{1}{n+1})a_n$ or $a_{n+1} = (1-\frac{1}{(n+1)^2})a_n$.

Recall that at $n = 0$, $a_{n+1} = 0$ because $(1-1)a_0 = 0$. Subsequent terms also equal zero (can you see why)? Your equation wouldn't hold because the ratio is undefined.

I will stop commenting now. If the above bothers you because of the $n=0$ case, replace $n+1$ by $n+2$ in the ratio above. But please, spend some time thinking about what I wrote: basing your answer on only the fact that $\lvert a_{n+1} \rvert < \lvert a_{n} \rvert$ (which you do) instead of $\lvert a_{n+1} \rvert < r\lvert a_{n} \rvert$ for $r < 1$ (which is the stronger, correct assumption to use) cannot work, since there are sequences satisfying the former such that $\sum_n a_n$ diverges to infinity.

Clearly, just take the harmonic series. But I never base my answer on that. Can't you see that I've included the ratio in my inequality? Are you legally blind?

You've stopped commenting because you know I'm right. I've invited you to a chat room where we can discuss this and hopefully sort out some of your legally blind issues.

Your answer (b)-(c) does not give a formal argument. Your argument for concavity above in the comments is handwavy at best, and when I base a counterexample on the claim you make "concave down", etc), you don't listen. Then you become rude. Goodbye.