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# Prove some theorems on convergence of series

Consider two sequences and whose terms are always positive after some integer (i.e., and for all ). Then define Prove the following two statements.

1. If there exists a constant with for all , then converges.
2. If the series diverges and if for all then the series diverges.

1. Proof. First, we solve the given relation for (starting with the term so that all terms are positive, hence, nonzero), Then, since for all , we have Therefore, the partial sums of are bounded (since is a constant); hence, converges 2. Proof. We are given which implies This then implies Since is a constant this implies that if diverges so does 1. Anonymous says:

Here is my proof for part a).

we know from the problem statement multiplying by and dividing by we get summing up such inequalities for to we have the first sum collapses by the telescopic property since we have the sequence of partial sums of is thus bounded. And since it’s increasing too ( ), then it must converge by theorem 10.1.

• I solved it the same way! I will add my proof of (b) :
$$b_n－\frac {b_{n+1}a_{n＋1}}{a_n}≤0$$ $$a_nb_n≤b_{n+1}a_{n＋1}$$
$$0≤a_Nb_N≤a_nb_n$$
$$0≤\frac {a_Nb_N}{b_n}≤a_n$$

2. Anonymous says:

Just noticed a couple of typos. See suggestions below.

In the proof for part a:
\begin{frame}
Then, since for all , we have
\end{frame}

In the proof for part b:
\begin{frame} \end{frame}

3. Anonymous says:

In the proof for part a, should the sentence “Then, since 0 < c_N \leq r for all n \geq N, we have" be "Then, since 0 < c_N \leq r for all n \geq N, we have" instead?

In the proof for part b, 