How do you find the sum of the infinite series #Sigma(1/10)^k# from k=1 to #oo#?

Question

Elijah Abram · Accepted Answer

#color(blue)(1/9)#

#sum _(k=1)^(oo)(1/10)^k#

Find the common ratio, by calculating the first three terms.

#(1/10)^1, (1/10)^2, (1/10)^3= 1/10, 1/100, 1/1000#

Ratio:

#(1/100)/(1/10)=(1/1000)/(1/100)=1/10#

This could have been seen from the summation expression.

The sum of a geometric series is:

#a((1-r^n)/(1-r))#

Where #a# is the first term, #r# is the common ratio and #n# is the nth term.

So:

#1/10((1-0)/(1-(1/10)))=1/10(1/(1/(9/10)))=1/10(10/9)=color(blue)(1/9)#

This could also have been arrived at using the limit to #oo#

#1/10lim_(n->oo)((1-(1/10)^n)/(1-(1/10)))=10/9=1/10*10/9=1/9#

Elijah Abram · Answer

undefined To find the sum of the infinite series Σ(1/10)^k from k=1 to ∞, you can use the formula for the sum of an infinite geometric series, which is given by S = a / (1 - r), where "a" is the first term and "r" is the common ratio. In this series, the first term "a" is 1/10 and the common ratio "r" is also 1/10. So, plugging these values into the formula, you get:

S = (1/10) / (1 - 1/10)

Simplify the expression:

S = (1/10) / (9/10)
S = (1/10) * (10/9)
S = 1/9

Therefore, the sum of the infinite series Σ(1/10)^k from k=1 to ∞ is 1/9.