The **root test** checks whether the **n ^{th}** root of the

The integral is a simple one (I've flipped the exponent and denominator of the integral from **-p+1** to **1-p** for convenience).

**R→∞**. With the p-integral in hand, we can use the integral test to determine which p-series converge.

Now generally we'll get a single fraction by recognizing that division is multiplication by the reciprocal:

Now we can use the properties of factorials and the laws of exponents to do some simplification. Recall that **2 ^{n+1} = 2^{n} ยท 2^{1}** and

which reduces to

The limit is less than one, so the series converges. Hopefully, you had a pretty good suspicion that this series was convergent from the beginning, because the factorial function grows much more rapidly than an exponential function for **n > N**, some number at or beyond the point where the two functions cross.

Remember that this test, and any convergence test doesn't necessarily tell us *which* number the series converges to, just that it does. We would need to find the *sum* of the series in another way.

Now, recognizing that **e ^{n+1}/e^{n} = e^{n}e^{1}/e^{n} = e**, and using L'Hopital's rule to evaluate the limit, we get:

The series diverges. Notice that this series wouldn't have passed the divergence test, either.

Now expand the **(n+1) ^{2}** and rearrange to keep the roots together:

The properties of limits allow us to separate this expression into a product of limits:

The properties of limits also allow us to put the power of 1/2 on the left outside of the limit ...

... to obtain a limit of **1**, which means that the ratio test is inconclusive for this series.

This series can be shown to be convergent by direct comparison to the p-series with **p = 3/2**.

Now we can use L'Hopital's rule to find the limit:

... so the series converges.

Because 0 < 1, the series converges. Notice that the series with terms -3/n would not converge (-3 times an arithmetic series, but that's not what's going on here).

Now the **n ^{3/n}** can be rewritten as

... so the series converges.

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