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Prove that:

\$\$limlimits_{n to infty} frac{1}{sqrt[4]{n}} = 0\$\$

Using the formal definition of convergence.

Workings:

\$frac{1}{sqrt[5]{n}} > frac{1}{sqrt[4]{n}}\$

To make \$frac{1}{sqrt[5]{n}}\$ less than \$epsilon\$.

Need \$frac{1}{sqrt[5]{n}}\$ < \$epsilon\$

That is \$frac{1}{sqrt[5]{epsilon}} < n\$

So let \$N = frac{1}{sqrt[5]{n}}\$

Proof:

Suppose \$epsilon > 0\$

Let \$N = frac{1}{sqrt[5]{n}}\$

For any \$n > N\$ we have that \$n > frac{1}{sqrt[5]{epsilon}}\$.

And we have:

\$|s_n – L| = left|frac{1}{sqrt[4]{n}} – 0right|\$

\$|s_n – L| = left|frac{1}{sqrt[4]{n}}right|\$

\$|s_n – L| = frac{1}{sqrt[4]{n}}\$

\$leq frac{1}{sqrt[5]{n}}\$

\$< frac{1}{N}\$

\$leq frac{1}{frac{1}{epsilon}}\$

\$= epsilon\$

Therefore:

\$\$limlimits_{n to infty} frac{1}{sqrt[4]{n}} = 0\$\$

I’m not sure if I did this correctly. Any help will be appreciated.