Divergence of series $\sum \frac{k}{2k+1}$

Question

Consider the infinite series denoted by: $$\sum \frac{k}{2k+1}$$ starting at some arbitrary index. We wish to prove that this diverges. We will see what the limit of the general term $a_n$ is, if this does not go to zero, the series diverges by the limit test. $$\lim_{k \rightarrow \infty} \frac{k}{2k+1}=\lim_{k \rightarrow \infty} \frac{1}{2+\frac{1}{k}}= \frac{1}{2+0}=\frac{1}{2}$$ By using the fraction limit theorem $(\lim \frac{A}{B}= \frac{\lim A}{\lim B})$ and sum limit theorem respectively. We conclude that the series diverges as $a_n \not \rightarrow 0$.

Is this correct? I wanted to also try the following proof for $k\geq 1$:

$$ \frac{k}{2k+1} > \frac{1}{2k+1} \geq \frac{1}{2k+k} = \frac{1}{3} \frac{1}{k} $$ Now we simply get the harmonic series, but every factor is multiplied by $\frac{1}{3}$, is there still some result we can use about multiplying divergent series by nonzero nonnegative constants?

Answer 1

Yes you are correct since $a_n \not \to 0$ the series does not converges and since it is with positive terms it diverges to $+\infty$.

Also the alternative way is fine.

Answer 2

Or you can use the following $${k\over 2k+1}\ge {k\over 3k}={1\over 3}$$for $k\ge 1$ which implies a direct divergence. Still, your proof sounds right to me.