The series #sum_(n=1)^oo x^n/10^n # converges for #|x| lt beta#, find #beta#?

Answer 1

#beta=10#

We can apply d'Alembert's ratio test:

Suppose that;

# S=sum_(n=1)^oo a_n \ \ #, and #\ \ L=lim_(n rarr oo) |a_(n+1)/a_n| #

Then

if L < 1 then the series converges absolutely; if L > 1 then the series is divergent; if L = 1 or the limit fails to exist the test is inconclusive.

Our series is;

# S = sum_(n=1)^oo a_n # with # a_n=x^n/10^n #

So our test limit is:

# L = lim_(n rarr oo) | ( x^(n+1)/10^(n+1) ) / ( x^n/10^n) | # # \ \ \ = lim_(n rarr oo) | ( x^(n+1)/10^(n+1) ) * ( 10^n/x^n ) | # # \ \ \ = lim_(n rarr oo) | ( (x x^n)/(10 * 10^n) ) * ( 10^n/x^n ) | # # \ \ \ = lim_(n rarr oo) | x/10 | # # \ \ \ = | x/10 | #
Comparing with the definition of the question, we can conclude that the series converges if #L lt 1#
# | x/10 | lt 1 => |x| lt 10 #
Similarly, the series diverges if #L gt 1#
# | x/10 | gt 1 => |x| gt 10 #
Hence, #beta=10#
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Answer 2

The series (\sum_{n=1}^{\infty} \frac{x^n}{10^n}) converges if (|x| < 10). Therefore, (\beta = 10).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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