If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Integral Calculus>Unit 5

Lesson 2: Infinite geometric series

# Convergent & divergent geometric series (with manipulation)

Sal looks at examples of three infinite geometric series and determines if each of them converges or diverges. To do that, he needs to manipulate the expressions to find the common ratio. Created by Sal Khan.

## Want to join the conversation?

• does a sequence have to converge if n approaches infinity if it's series converges or are they mutually exclusive?
• They can both converge or both diverge or the sequence can converge while the series diverge.

For example, the sequence as n→∞ of n^(1/n) converges to 1 . However, the series
∑ n=1 to ∞ n^(1/n) diverges toward infinity.

As far as I know, and I might be wrong about this (but I am fairly sure) that a sequence must converge in order for its series to converge (though the series is not certain to converge just because the sequence converges). So, yes, IF a series converges as n→∞, then its corresponding sequence must also converge.
• How do we find what the value converges to?
• @Arsenio Trucco note that that formula will only work for some rⁿ when n starts as zero. The sum from n=0 to infinity of a series is not always the same as the sum from n=5 to infinity of that series, because the first few terms are not counted towards the sum. You can compensate for this by using the proof in previous videos to discover that given that n starts at a constant b, Sn-rSn=ar^b, so Sn = (ar^b)/(1-r). This edited formula works for n=0 too, because ar^0 is just a, like the previous formula
• Could someone tell me what is the difference between a series and a sequence ?
• A sequence is a list of numbers in a specific order. A series is the result of adding the terms of a sequence.
• from to how did he move from 2^(n)/3^(n)3^(-1) to 3* 2^(n)/3^(n) ? Did he just multiply by 3 to cancel the 3^(-1) ?
• Well, 1 / (3^(-1)) = 1 / (1 / 3) = 1 * 3 / 1 = 3. So:

(2^n) / (3^n 3^(-1)) =
(2^n) / (3^n) * 1 / (3^(-1)) =
(2^n) / (3^n) * 1 / (1 / 3) =
(2^n) / (3^n) * 3 =
3(2^n) / (3^n)

Write it out on paper, without parentheses, and it may become clear.
• I have a question. In my math class I was told that geometric series always had to start from n=0 or n=1, or at least that is what I understood when I went to class. However, Sal treats every series that has a common ratio as a geometric series regardless of the number at which n starts. So, is it possible that geometric series start at any number??
• Around , what was the mental math Sal did with the fractions? I understand that he was actually multiplying 1*3 and 2*9 and simplifying, but he muttered something about, "3 over 9... so we're gonna get 1." If he was using some kind of shorthand trick for multiplying fractions, that's a trick I'd love to know.
• I can't read his mind, but my guess would be: 3/2 * 1/9 = 3/9 * 1/2 = 1/3 * 1/2 = 1/6. In other words, he immediately saw that 3 and 9 are "related": 9 is divisible by 3, so if you have those on opposite sides of the fraction line (one above, one below) then you can simplify the fraction by dividing both by 3.
(1 vote)
• if an infinite series is neither geometric nor arithmetic ,...e.g. 1+2x+3x^2+4x^3..... then please suggest some way to solve such problems
(1 vote)
• solve a polynomial? what do you mean by solve
• Man I am so confused. Is there a reason he is using algebra to purposely arrive at only one n exponent?

For example, his first equation. (5^(n-1))(9/10)^n. Why can't we just immediately apply the limit as n approaches infinity?

(9/10)^n as n approaches infinity equals zero.
5^(n-1) as n approaches infinity equals infinity.

Infinity (zero) = converges at zero. But obviously that is not what he got. So confused.
(1 vote)
• Remember that 0 and ∞ are approached, never equaled: so the rule that 0*anything = 0 does not apply when multiplied by ∞ because you have two rules in conflict. ∞ times anything approaches infinity while 0*anything approaches 0; thus these two rules conflict and the answer is indeterminate -- that is, the rules don't tell us what the limit is.
• Is it not easier to find r by simply dividing any a(n) by a(n-1)?
(1 vote)
• at , why do you exclude 1/5 from the ratio? Doesn't it have to be (5*9)^n/(5*10)?
(1 vote)
• Hi,

That's because he applies one of the laws of exponents where:

aˣ * bⁿ * cⁿ = aˣ * (bc)ⁿ.

It's possible to gather numbers or variables under a common exponent if they are raised to the same exponent. Since 1/5 is not raised to the nth power, it was excluded, yielding the value of 'a' in a * rⁿ. (the 10 in the denominator is raised to the nth power as well).