I am struggling with convergence and limits. In theory, I know the ratio test and comparison test for absolute convergence (vaguely aware of the integral test), and Leibnitz's criterion for alternating series. However when it comes to using one of these, it is much harder. The ratio test in particular gives me some difficulty, as it seems you need to have some awareness of limits to use it anyway.

For example, I have the series formed by the terms $(\frac { 2n+5}{3n+1})^n $ In trying the ratio test on it to find if it converges, I came to

$lim_{n->\infty} (\frac { 2n+7}{3n+4})^{n+1}(\frac {3n+1}{2n+5})^n$

Now I am having trouble as the first bracket tends to 0 whereas the second tends to infinity. How would I approach such a limit? Does it require more applications of the ratio test or perhaps the ratio test has simple 'failed' here?

  • 1
    $\begingroup$ Before using a test, whether the ratio test or another one, try to guess what the sequence looks like. Here, for every $n\geqslant5$, $$\frac{2n+5}{3n+1}\leqslant\frac{15}{16}$$ hence $$\left(\frac{2n+5}{3n+1}\right)^n\leqslant r^n\qquad r=\frac{15}{16}<1$$ which shows the convergence of the series, by comparison with the converging series $$\sum_nr^n$$ $\endgroup$ – Did Jan 12 '17 at 13:50

The useful criteria here is comparison with a geometric series.

One can check that, for $n \in \mathbb{N}$,

$$n>17 \ \ \ \iff \ \ \ \dfrac{2n+5}{3n+1}<\dfrac{3}{4}$$

Thus $$\sum_{n=18}^{\infty}\left(\dfrac{2n+5}{3n+1}\right)^n<\sum_{n=18}^{\infty}\left(\dfrac{3}{4}\right)^n<\sum_{n=0}^{\infty}\left(\dfrac{3}{4}\right)^n=\dfrac{1}{1-\frac34}=4.$$

proving convergence of


because adding (or removing) a finite number of terms to a series does not modify its nature (convergent or divergent).


When a term looks like $\frac{an+b}{cn+d}$, intuitively we might think that for sufficiently high $n$, the constant terms will be negligible and can be ignored.

To get a more rigorous justification for this, we let $n>N$ and fix $\varepsilon = 1/N$. Then we can see that $$\frac{2n}{3n+\varepsilon n} < \frac{2n+5}{3n + 1} < \frac{2n+5 \varepsilon n}{3n}.$$

At this point we can use the comparison test and let $\varepsilon\to 0$ to show this series converges; you could use a similar technique to show that the sequence produced in the ratio test converges to $\frac{2}{3}$.

In general when faced by the limiting behaviour of this sort of rational function, it often helps to reduce the numerator and denominator to multiples of their respective dominating terms by some method like the above.


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