# Ratio test for series with eventually positive coefficients

We have a generic real series $$\sum_{n=0}^{+\infty} a_n$$, where $$a_n>0$$ for all $$n$$.

If we have $$\frac{a_{n+1}}{a_n} \to l<1$$ then $$\sum_{n=0}^{+\infty} a_n$$ converges, while if $$l>1$$ it diverges.

Now, suppose that $$a_n=0$$ if $$n \le n_0$$ and $$a_n>0$$ if $$n>n_0$$.

In this case, if $$\frac{a_{n+1}}{a_n} \to l<1$$ (where obviously $$n>n_0$$) can we still conclude that $$\sum_{n=0}^{+\infty} a_n$$ converges?

• Finite amount of elements never affects on the divergence or convergence of a series/sequence. – eminem Sep 16 at 12:16
• So the answer is yes, right? Thank you! – Leonardo Sep 16 at 12:18

Yes, you can still conclude that the series is convergent.

Define $$b_k = a_{(k + n_0 + 1)}.$$

From your previous statements, $$\sum_{n=0}^{+\infty} b_n$$ is a convergent series.

But this is the same as $$\sum_{n=0}^{+\infty} a_n,$$ except for a finite # of terms that are at the start of $$\sum_{n=0}^{+\infty} a_n.$$

Therefore, both series converge or they both diverge.

Therefore, since $$\sum_{n=0}^{+\infty} b_n$$ is a convergent series
so is $$\sum_{n=0}^{+\infty} a_n$$.

In general for any series with $$a_n$$ well defined $$\forall n\ge n_0$$

$$\sum_{n=n_0}^{\infty} a_n<\infty \iff \sum_{n=n_1}^{\infty} a_n<\infty$$

that is the series convergence (or divergence) is determinated by the tails and not by any finite number of initial terms, indeed

$$\sum_{n=n_0}^{N} a_n=\sum_{n=n_0}^{n_1-1} a_n+\sum_{n=n_1}^{N} a_n=S_{(n_1-1)}+\sum_{n=n_1}^{N} a_n$$

and

$$\sum_{n=n_0}^{\infty} a_n=L \iff \lim_{N\to \infty} \sum_{n=n_0}^{N} a_n=L \iff \lim_{N\to \infty} \sum_{n=n_1}^{N} a_n=L-S_{(n_1-1)}$$

and since