Decide whether the infinite series converges of diverges. $$\sum_{n=1}^\infty \frac{2^n+3^n}{3^n+4^n}$$

My thought process:
The nth term test doesn't seem viable after initial use.
The comparison function I derived for DCT/LCT is: $\frac{3^n}{4^n}$. I cannot create an appropriate inequality between the given function and my comparison function I found, so I tried LCT. I am aware that $\sum_{i=1}^\infty \frac{3^n}{4^n}$ is geometric and because the common ratio (r) is $ \frac{3}{4}$, $-1<r<1$ the series converges, however I cannot compute the limit LCT requires.
Application of the root test would seem unbeneficial because roots do not split over sums.
As for the ratio test I was also unsuccessful in proceeding with the computation of the limit.

The above are the only tests I have learned as of right now. My question is what test should I be looking to utilize to determine convergence or divergence?

  • 2
    $\begingroup$ $2^n +3^n < 2\times 3^n$, $3^n +4^n > 4^n$. $\endgroup$
    – Fnacool
    Apr 13 '17 at 3:24
  • 1
    $\begingroup$ When you comeback in this site, please read this $\endgroup$
    – Juniven
    Apr 13 '17 at 8:02

Use the comparison test.

For every $n$, let $$ a_n = \dfrac{2^n + 3^n}{3^n + 4^n}. $$ The given series is $\sum_{n=1}^{\infty} a_n$. We will show that it converges.

For every $n$, $$ a_n = \dfrac{2^n + 3^n}{3^n + 4^n} \leq \dfrac{3^n + 3^n}{4^n} = 2 \biggl( \dfrac{3}{4} \biggr)^n. $$ For every $n$, let $b_n = 2(\tfrac{3}{4})^n$. So $a_n \leq b_n$ for every $n$. Since $\sum_{n=1}^{\infty} (\tfrac{3}{4})^n$ is a geometric series with common ratio $\tfrac{3}{4}$, it converges. Thus, $2 \sum_{n=1}^{\infty} (\tfrac{3}{4})^n$ converges also, i.e., $\sum_{n=1}^{\infty} b_n$ converges. Therefore, $\sum_{n=1}^{\infty} a_n$ converges by the comparison test.


Let us try to split:


Then we consider each the following series:

$$\sum_{n=1}^\infty \frac{2^n}{3^n+4^n}\quad\text{and}\quad \sum_{n=1}^\infty \frac{3^n}{3^n+4^n}\tag 1$$

As we can see, we have $$\frac{2^n}{3^n+4^n}<\frac{2^n}{3^n}=\bigg(\frac{2}{3}\bigg)^n\quad \forall n\in\Bbb N$$ and $$\frac{3^n}{3^n+4^n}<\frac{3^n}{4^n}=\bigg(\frac{3}{4}\bigg)^n\quad \forall n\in\Bbb N$$

Convergence of the given two series in $(1)$ follows from the Comparison Test because
$$\sum_{n=1}^\infty \bigg(\frac{2}{3}\bigg)^n\quad\text{and}\quad \sum_{n=1}^\infty \bigg(\frac{3}{4}\bigg)^n$$ are both convergent. Hence, the series $$\sum_{n=1}^\infty \frac{2^n+3^n}{3^n+4^n}$$ is convergent (being a sum of two convergent series).


It's enough to notice that all terms $a_n$ of your sequence satisfy $$ 0< a_n < q^n \tag{1} $$ for an appropriately chosen $q<1$. For example, you can choose $q=0.9$, and $(1)$ will be true for all $n\ge1$. Then you can use the fact that the geometric series $$ \sum_{n=1}^\infty q^n = {q\over 1-q} $$ converges; therefore your series $\sum_{n=1}^\infty a_n$ converges too, by comparison test.


Besides the already provided solutions, I'd like to point out that both the LCT and the Root Test do work here too.

If we're applying the LCT to $\sum_{n=1}^\infty \frac{2^n+3^n}{3^n+4^n}$ and $\sum_{n=1}^\infty \frac{3^n}{4^n}=\sum_{n=1}^\infty \left(\frac{3}{4}\right)^n$:


so the LCT applies to these two series and they behave the same way.

A similar trick works with the Root Test too. We can factor our $3^n$ and $4^n$ in the numerator and the denominator, respectively, to find that


UPDATE: I've just noticed that the OP mentioned the Ratio Test too. I'm not going to type one more solution, but the limit of the fraction in the Ratio Test can be simplified here using a trick similar to the one used in the LCT calculation above.

  • $\begingroup$ Very useful. I completely overlooked some of the manipulations/tricks you utilized. Cheers! $\endgroup$
    – Wheat
    Apr 13 '17 at 4:46

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