Question

If \(\sum {{{\rm{c}}_{\rm{n}}}{{\rm{6}}^{\rm{n}}}} \)is convergent, then\(\sum {{{\rm{c}}_{\rm{n}}}{{{\rm{( - 2)}}}^{\rm{n}}}} \)is convergent.

Short answer

Given statement is true; If \(\sum {{c_n}{6^n}} \)is convergent, then\(\sum {{c_n}{{( - 2)}^n}} \)is convergent.

Step-by-step solution

Test of convergence and divergence

If the sequence\({{\rm{S}}_{\rm{n}}}\)is convergent and\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{S}}_{\rm{n}}}{\rm{ = s}}\)exists as a real number; then the series\(\sum {{{\rm{S}}_{\rm{n}}}} \)is convergent.

If\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{S}}_{\rm{n}}}\)does not exist or if\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{S}}_{\rm{n}}} \ne {\rm{0}}\), then the series\(\sum {{{\rm{S}}_{\rm{n}}}} \)is divergent.

Step 2:

Given series \(\sum {{c_n}{6^n}} \)is convergent.

i.e., \(\mathop {\lim }\limits_{n \to \infty } {c_n}{6^n} = 0\)

\({c_n}\)should be less than\(\frac{1}{{{6^n}}}\)for the limit to be 0.

In other words, denominator of\({c_n}\)as a fraction would be more than\({6^n}\).

For \(\sum {{c_n}{{( - 2)}^n}} \),

\(\mathop {\lim }\limits_{n \to \infty } {c_n}{2^n} = 0\)

Therefore, by alternating series test If \(\sum {{c_n}{6^n}} \)is convergent, then\(\sum {{c_n}{{( - 2)}^n}} \)is convergent.