# Infinite Series convergence

• fender5791
That is incorrect. Try substituting a few values of k and you should immediately see what this term is. In general we have the rule:\frac{a^x}{a^y} = a^{x-y}which you can use to calculate this term.Anyway if you want to show that a series \sum a_i diverges a common way is to show that you can find a positive number M such that a_i \geq M. In your specific example can you find a lower bound for 1-\frac{2^{k-1}-1}{2^{k+1}}?

#### fender5791

Okay, there's two questions, actually.

First, determine if the series converges.

SUM: (n-2)/(n^2-4n) (from n=5 to infinity)

I used the integral test, found the integral to be 1/2 log(n^2-4n) from x=5 to x=t as t approaches infinity. That turned out to go to infinity so the series diverges.

Does this seem right or did I do something wrong?

Second, Suppose a series is the SUM: (1 - ((2^(k-1) - 1) / (2^(k+1)))

Does the sum of the series converge or diverge? I'm not really even sure where to start this one.

fender5791 said:
Okay, there's two questions, actually.

First, determine if the series converges.

SUM: (n-2)/(n^2-4n) (from n=5 to infinity)

I used the integral test, found the integral to be 1/2 log(n^2-4n) from x=5 to x=t as t approaches infinity. That turned out to go to infinity so the series diverges.

Does this seem right or did I do something wrong?
I used the ratio test and concluded the same thing.
fender5791 said:
Second, Suppose a series is the SUM: (1 - ((2^(k-1) - 1) / (2^(k+1)))

Does the sum of the series converge or diverge? I'm not really even sure where to start this one.

Here's what I think your series is:
$$\sum_{k = 1}^{\infty} \frac{1 - 2^{k - 1} - 1}{2^{k + 1}}$$

If that's what it is, there's a lot of simplification you can do. After that you should be able to quickly determine whether it converges or not.

Mark44 said:
I used the ratio test and concluded the same thing.

Here's what I think your series is:
$$\sum_{k = 1}^{\infty} \frac{1 - 2^{k - 1} - 1}{2^{k + 1}}$$

If that's what it is, there's a lot of simplification you can do. After that you should be able to quickly determine whether it converges or not.

No sorry, the first one is outside, then subtract the quantity starting with 2^k-1

Like this?
$$\sum_{k = 1}^{\infty} \left(1 - \frac{2^{k - 1} - 1}{2^{k + 1}}\right)$$

Try expanding the series to see what the first few terms look like.

Mark44 said:
Like this?
$$\sum_{k = 1}^{\infty} \left(1 - \frac{2^{k - 1} - 1}{2^{k + 1}}\right)$$

Try expanding the series to see what the first few terms look like.

YEs, and I tried that. I did the limit as k went to infinity of 2^(k-1) / 2^(k+1) which I think is one, then the last bit is zero, so the total went to zero which doesn't help me much.

fender5791 said:
YEs, and I tried that. I did the limit as k went to infinity of 2^(k-1) / 2^(k+1) which I think is one, then the last bit is zero, so the total went to zero which doesn't help me much.

That is incorrect. Try substituting a few values of k and you should immediately see what this term is. In general we have the rule:
$$\frac{a^x}{a^y} = a^{x-y}$$
which you can use to calculate this term.

Anyway if you want to show that a series $\sum a_i$ diverges a common way is to show that you can find a positive number M such that $a_i \geq M$. In your specific example can you find a lower bound for
$$1-\frac{2^{k-1}-1}{2^{k+1}}$$
?

Thank you, got it now.