Home > Calculus > Integral Test for Convergence of an Infinite Series

Why does the integral test not apply to #Sigma (2+sinn)/n# from #[1,oo)#?

Answer 1

Because the condition only applies if #f(x)# and #a_n=f(n)# are non-negative and monotone decreasing, which is not the case here.

In fact, if we choose:

#f(x) = (2+sinx)/x#

we have that the function is continuous and non-negative for #x>0#, but as:

#f'(x) = frac(xcosx-2-sinx) (x^2)#

is positive for some values of #x#, clearly #f(x)# is not monotone decreasing.

graph{(2+sinx)/x [-10, 10, -5, 5]}

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Answer 2

Luke Alvarez

The integral test cannot be applied to ( \sum \frac{2+\sin(n)}{n} ) from ( n = 1 ) to ( \infty ) because the function ( f(n) = \frac{2+\sin(n)}{n} ) is not continuous, positive, and decreasing for all ( n \geq 1 ), which are the necessary conditions for the integral test to be applicable.

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Answer 3

Emilia Aguilar

The integral test cannot be directly applied to the series Σ(2 + sin(n))/n from [1, ∞) because the function (2 + sin(n))/n is not integrable over the interval [1, ∞). This means that the corresponding improper integral does not converge, and thus, the integral test cannot be used to determine the convergence of the series.

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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.