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?

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    $\begingroup$ Finite amount of elements never affects on the divergence or convergence of a series/sequence. $\endgroup$ – eminem Sep 16 at 12:16
  • $\begingroup$ So the answer is yes, right? Thank you! $\endgroup$ – 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 $\sum_{n=0}^{+\infty} a_n$.

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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

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


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

and since

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