Does the series with increasing numerators converge?

[karush]

Find the sum for the series

$$\frac{5}{3}+2+\frac{12}{5}+...$$
This equals

$$\frac{25}{15}+\frac{30}{15}+\frac{36}{15}+...$$

So the numerator increases by 4+k from the previous numerator
But unable to set up
$$\sum_{k+1}^{\infty}f(x)$$
The series should go to $\infty$ since the terms only increase in size

 

[suluclac]

Correct, the series diverges.

 

[MarkFL]

Find the sum for the series

$$\frac{5}{3}+2+\frac{12}{5}+...$$
This equals

$$\frac{25}{15}+\frac{30}{15}+\frac{36}{15}+...$$

So the numerator increases by 4+k from the previous numerator
But unable to set up
$$\sum_{k+1}^{\infty}f(x)$$
The series should go to $\infty$ since the terms only increase in size

Given that the numerator increases linearly, we know it will be a quadratic function and so the $n$th term in the series will have the form:

$$a_n=\frac{an^2+bn+c}{15}$$

Using the given initial values, the following 3X3 linear system arises:

$$a+b+c=25$$

$$4a+2b+c=30$$

$$9a+3b+c=36$$

Solving this system, we obtain:

$$(a,b,c)=\left(\frac{1}{2},\frac{7}{2},21\right)$$

And so we have the $n$th term:

$$a_n=\frac{n^2+7n+42}{30}$$

Hence, the $n$th sum of the series is given by:

$$S_n=\sum_{k=1}^n\left(a_k\right)=\frac{1}{30}\left(\sum_{k=1}^n\left(k^2+7k+42\right)\right)=\frac{1}{30}\left(\sum_{k=1}^n\left(k^2\right)+7\sum_{k=1}^n(k)+42\sum_{k=1}^n(1)\right)$$

Now, using the following formulas:

$$\sum_{k=1}^n\left(k^2\right)=\frac{n(n+1)(2n+1)}{6}$$

$$\sum_{k=1}^n\left(k\right)=\frac{n(n+1)}{2}$$

$$\sum_{k=1}^n\left(1\right)=n$$

We may now state:

$$S_n=\frac{1}{30}\left(\frac{n(n+1)(2n+1)}{6}+\frac{7n(n+1)}{2}+42n\right)=\frac{n}{180}\left((n+1)(2n+1)+21(n+1)+252\right)$$

Expanding, combining like terms and factoring, we finally obtain:

$$S_n=\frac{n}{90}\left(n^2+12n+137\right)$$

And so we find:

$$S_{\infty}=\lim_{n\to\infty}\left(S_n\right)=\infty$$

 

[soroban]

Find the sum for the series: \frac{5}{3}+2+\frac{12}{5}+ \cdots
This equals: \frac{25}{15}+\frac{30}{15}+\frac{36}{15}+\cdots
So the numerator increases by 4+k from the previous numerator , but unable to set up \sum_{k+1}^{\infty}f(x)
The series should go to $\infty$ since the terms only increase in size

I saw a different series.

\frac{5}{3} + 2 + \frac{12}{5} + \cdots \;=\;\frac{5}{3} + \frac{8}{4} + \frac{12}{5} + \cdots \;=\;\sum^{\infty}_{n=1} \frac{n^2+3n+6}{2(n+2)} . . . which also diverges.

 

[HallsofIvy]

You should see immediately that the individual terms do not go to 0 so the series does not converge.