anyone good at Infinite Geometric Series?

[danabear16]

I'm having trouble with the Infinite Geometric Series with the infinity symbol over the sigma. I don't really know what the type of problem is called.
∞
Σ (5/3)^k
k=1

any help would be greatly appreciated. thanks

 

[Unco]

An infinite geometric series diverges if the common ratio, here 5/3, is greater than or equal to 1.

 

[soroban]

Hello, danabear16!

I'm having trouble with the Infinite Geometric Series with the infinity symbol over the sigma.
I don't really know what the type of problem is called.

\(\displaystyle \L\sum^{\infty}_{k=1}\left(\frac{5}{3}\right)^k\)

You already called it . . . it's an infinite geomtric series, the terms go on forever.
. . The sum may be infinite, of course, but sometimes it has a finite sum.

If you write out the summation: \(\displaystyle \:\frac{5}{3}\,+\,\left(\frac{5}{3}\right)^2\,+\,\left(\frac{5}{3}\right)^3\,+\,\left(\frac{5}{3}\right)\,+\,\cdots\)

. . you'll see that it's a geometric series with first term \(\displaystyle a = \frac{5}{3}\) and common ratio \(\displaystyle r = \frac{5}{3}\)

An infinite geometric series converges if \(\displaystyle |r|\,<\,1\)
. . that is, \(\displaystyle r\) is between -1 and +1.

Your series has \(\displaystyle r\,=\,\frac{5}{3}\,>\,1\), so as Unco pointed out, it diverges.

 

[danabear16]

thank you so much soroban. that helped a lot.

 

[danabear16]

Hello, danabear16!

I'm having trouble with the Infinite Geometric Series with the infinity symbol over the sigma.
I don't really know what the type of problem is called.

\(\displaystyle \L\sum^{\infty}_{k=1}\left(\frac{5}{3}\right)^k\)

You already called it . . . it's an infinite geomtric series, the terms go on forever.
. . The sum may be infinite, of course, but sometimes it has a finite sum.

If you write out the summation: \(\displaystyle \:\frac{5}{3}\,+\,\left(\frac{5}{3}\right)^2\,+\,\left(\frac{5}{3}\right)^3\,+\,\left(\frac{5}{3}\right)\,+\,\cdots\)

. . you'll see that it's a geometric series with first term \(\displaystyle a = \frac{5}{3}\) and common ratio \(\displaystyle r = \frac{5}{3}\)

An infinite geometric series converges if \(\displaystyle |r|\,<\,1\)
. . that is, \(\displaystyle r\) is between -1 and +1.

Your series has \(\displaystyle r\,=\,\frac{5}{3}\,>\,1\), so as Unco pointed out, it diverges.

one more question... what do I do if K=0?

 

[Gene]

You mean
\(\displaystyle \L\sum^{\infty}_{k=0}\left(\frac{5}{3}\right)^k\)

Well (5/3)^0 = 1 so it = \(\displaystyle {\infty}+1\)
I doubt if you would notice the difference :evil:

BTW it isn't necessary to copy an entire previous post. It's only a few clicks away.