Methods to prove a series is decreasing? What are soome good methods to show a series is decreasing?
Usually I just compare the $a_n$ and $a_{n+1}$ values. For example $\frac 1 {k^2} >\frac 1 {k^2+1}$ so it shows the series is decreasing. I heard you can also do something like $\frac{a_{n+1}}{a_n}$ but that seems complicated.
What are some more efficient methods for like series that have $\frac{k+2} {k(k+3)}$?
 A: If you are more comfortable with things like $\frac{1}{k^2} > \frac{1}{k^2 +1}$ then another approach for cases like $\frac{k+2}{k(k+3)}$ is to use partial fraction decomposition:
$$\frac{k+2}{k(k+3)} - \frac{k+3}{(k+1)(k + 4)} = \left(\frac{1/3}{k + 3} + \frac{2/3}{k}\right) - \left(\frac{1/3}{k + 4} + \frac{2/3}{k+1}\right) = \frac{1}{3}\left(\frac{1}{k+3} - \frac{1}{k+4}\right) + \frac{2}{3}\left(\frac{1}{k} - \frac{1}{k+1}\right)$$
A: HINT: Write the sequence in the form of a function, say $f(x)$ and to show that $f(x)$ is decreasing, show that $f'(x)<0$.
A: Yes, considering $\frac{a_{n+1}}{a_n}$ is a good bet, and is actually equivalent to your current approach: $a_{n+1}<a_n \Leftrightarrow \frac{a_{n+1}}{a_n}<1$ by simplying dividing both sides by $a_n$ (given that $a_n>0$).
When series' terms have polynomials, try dividing both the numerator and denominator by the highest power. For example, if you have $\frac{a_{k+1}}{a_k}=\frac{8x^3+2x^2}{3x^4+x}$, the highest power in the fraction is $x^4$. Notice that for $x\ne0$, $1=\frac{1/x^4}{1/x^4}$. So $\frac{8x^3+2x^2}{3x^4+x}=\frac{(8x^3+2x^2)/x^4}{(3x^4+x)/x^4}=\frac{8/x+2/x^2}{3+1/x^3}$. Then you can solve for when this term is less than 1.
A: if a series is monotonic ( increasing or decreasing) and it converges to L. then you just need to compare the first value and the limit. if the limit is bigger then the series is increasing. if L is less that the first value then the series is decresing. In your case the series converges to 0 and the first value when k=0 you have 3/4. so just prove that it is monotonic and it will follow that it is decreasing.
Good Luck
