# How do you find the interval of convergence of #Sigma (x+10)^n/(lnn)# from #n=[2,oo)#?

The series:

is convergent for

and the series is divergent.

and apply the Leibniz test:

and

so the series is convergent.

In conclusion the series is convergent for:

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To find the interval of convergence of the series Σ((x+10)^n / ln(n)) from n=2 to infinity, we can use the ratio test. According to the ratio test, the series converges if the absolute value of the ratio of successive terms approaches a finite number as n approaches infinity.

So, we calculate the limit as n approaches infinity of |((x+10)^(n+1) / ln(n+1)) / ((x+10)^n / ln(n))|.

This simplifies to the absolute value of ((x+10)(n) / (n+1)) * (ln(n) / ln(n+1)).

As n approaches infinity, ln(n) / ln(n+1) approaches 1, and (n / (n+1)) approaches 1.

Thus, the limit simplifies to |x + 10|.

For the series to converge, |x + 10| must be less than 1.

Therefore, the interval of convergence is -11 < x < -9.

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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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