Infinite series sum question?

Question

could anyone help with a relatively simple infinte series sum question I cant seem to get the answer to? (sigma n->inf, an= (3^n)/(4^(n-1)), the sum is 12, but by the equation a/1-r I keep getting 12/-23?

Answer 1

If an infinite series in SIGMA notation can be re-written as a GEOMETRIC series, in SIGMA notation, with the counting parameter starting with 1 and the exponent of the common ratio being 1 less than the counting parameter (as this one can), that is what you do to find the result (and it appears you are using "n" as your counting parameter).

SIGMA notation REQUIRES a starting value for the counting parameter. Your problem does not state the counting parameter's initial value. The final result of an infinite series will be different for a different initial counting parameter value.

To be correctly written with ALL relevant parameters AND get 12 as the answer, your problem should be written as

EITHER

Find the limit as n->infinity of sigma from k=1 to n of (3^k)/(4^(k-1))

OR

Find sigma from n=1 to infinity of (3^n)/(4^(n-1)

Notice that the "EITHER" option is written as a "limit" and the "OR" option does not use "limit" terminology (even though the concept of limits will be used in both options).

In the case of both options, factor out a 3 so that the series terms can be re-written as a geometric sequence:

EITHER

3*(3^(k-1))/(4^(k-1))=3((3/4)^(k-1)), from k to n,

with the common ratio 3/4 and k approaching infinity.

OR

3*(3^(n-1))/(4^(n-1))=3((3/4)^(n-1)), from n=1 to infinity,

with the common ratio 3/4

Your problem can NOW be written in SIGMA geometric series format:

EITHER

Find the limit as n->infinity of SIGMA from k=1 to n of 3((3/4)^(k-1))

with the common ratio 3/4 and k approaching infinity

OR

Find sigma from n=1 to infinity of 3((3/4)^(n-1))

with the common ratio 3/4

NOW, use the formula:

EITHER

the limit as n->infinity of SIGMA from k=1 to n of a(r^(k-1)) equals a/(1-r)

when the common ratio, r, is between, but not including, -1 and +1.

OR

sigma from n=1 to infinity of a(r^(n-1)) equals a/(1-r)

when the common ratio, r, is between, but not including, -1 and +1.

EITHER

The limit as n->infinity of SIGMA from k=1 to n of 3((3/4)^(k-1))=3/(1-3/4)

with the common ratio 3/4 being between, but not including, -1 and +1

OR

Sigma from n=1 to infinity of 3((3/4)^(n-1))

with the common ratio 3/4 being between, but not including, -1 and +1

AND

3/(1-3/4)=3/(1/4)=3*4=12

Answer 2

a₁ = 3

r = 3/4

S = a₁ / (1 - r)

S = 3 / (1/4)

S = 12

QED