Compute the sum of the given series. If the series diverges, enter DNE. Use exact values.
∑
n
=
3
∞
\displaystyle {\sum_{{{n}={3}}}^{\infty}}
n
=
3
∑
∞
(
−
1
)
n
−
1
2
n
−
2
8
n
=
\displaystyle \frac{{{\left(-{1}\right)}^{{{n}-{1}}}}}{{{2}^{{{n}-{2}}}{8}^{{n}}}}=
2
n
−
2
8
n
(
−
1
)
n
−
1
=
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\displaystyle
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)
Enter DNE for Does Not Exist, oo for Infinity