Question

Notify Me Bookmark

Find the sum of the series. summation n = 0 infinity 5^n/(6^n n!)

Find the sum of the series. summation n = 0 infinity 5^n/(6^n n!)

Image text
Find the sum of the series. summation n = 0 infinity 5^n/(6^n n!)

Detailed Answer

3 0

Please enter your email here. You will get email when this question is answered by our tutor.

Answer
Doubtrix Logo

This Problem has been solved!

You'll get a detailed, step-by-step and expert verified solution.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Login Sign Up
  • Student Reviews:
  • (3)
  • Correct answers (2)
  • Complete solution (2)
  • Step-by-step solution (2)
  • Fully explained (2)

Similar/Related Questions

This is a short reading exercise on reindexing. There is no standard name to this operation. Some people call it shifting the summation. Consider the notation for a series P∞ n=1 an. The symbol n is called the index of summation. The lower bound of the series is 1 . If we write P∞ n=u bn, then the lower bound of the series is u. Reindexing means to change lower bound of the series without changing the series. For example, the series 1 + 1 4 + 1 9 + 1 16 + 1 25 + • • • can be written as P∞ n=1 1 n2 or P∞ n=3 1 (n−2)2 . Please verify this by writing down the first few terms of both series. Therefore, we have an equality between series X∞ n=1 1 n2 = X∞ n=3 1 (n − 2)2 Notice we DO NOT simply replace n = 1 by n = 3. The expression for the terms of the series also have to change for the equality to hold. To reindex by symbolic manipulation, we introduce a change of variable. In this example, we start with the series P∞ n=1 1 n2 . We wish to reindex from n = 1 to n = 3, meaning the lower bound of the series changes from n = 1 to n = 3 without changing the series. We introduce m = n + 2 (because m = 3 when n = 1 ). Isolate for n to get n = m − 2, and we replace every occurrence of n in the series by the expression on the right-hand side (which is m − 2 in this example). The series becomes X∞ n=1 1 n2 = X∞ m−2=1 1 (m − 2)2 ( replace n by m − 2) = X∞ m=3 1 (m − 2)2 ( reorganize the lower bound of summation) = X∞ n=3 1 (n − 2)2 (change m back to n ) In general, if we wish to reindex a series P∞ n=u an from n = u to n = v, we introduce a change of variable m = n+(v−u) and perform similar manipulations as shown above. In the example above, u = 1 and v = 3. Let r be a non-zero constant. You are asked to reindex the series P∞ n=1 1 rn−1 from n = 1 to n = 0. What is the result of reindexing? Select one or more: a. P∞ n=0 1 rn+1 b. P∞ n=0 1 rn−1 c. P∞ n=0 1 rn

This is a short reading exercise on reindexing. There is no standard name to this operation. Some people call it shifting the summation. Consider the notation for a series P∞ n=1 an. The symbol n is called the index of summation. The lower bound of the series is 1 . If we write P∞ n=u bn, then the lower bound of the series is u. Reindexing means to change lower bound of the series without changing the series. For example, the series 1 + 1 4 + 1 9 + 1 16 + 1 25 + • • • can be written as P∞ n=1 1 n2 or P∞ n=3 1 (n−2)2 . Please verify this by writing down the first few terms of both series. Therefore, we have an equality between series X∞ n=1 1 n2 = X∞ n=3 1 (n − 2)2 Notice we DO NOT simply replace n = 1 by n = 3. The expression for the terms of the series also have to change for the equality to hold. To reindex by symbolic manipulation, we introduce a change of variable. In this example, we start with the series P∞ n=1 1 n2 . We wish to reindex from n = 1 to n = 3, meaning the lower bound of the series changes from n = 1 to n = 3 without changing the series. We introduce m = n + 2 (because m = 3 when n = 1 ). Isolate for n to get n = m − 2, and we replace every occurrence of n in the series by the expression on the right-hand side (which is m − 2 in this example). The series becomes X∞ n=1 1 n2 = X∞ m−2=1 1 (m − 2)2 ( replace n by m − 2) = X∞ m=3 1 (m − 2)2 ( reorganize the lower bound of summation) = X∞ n=3 1 (n − 2)2 (change m back to n ) In general, if we wish to reindex a series P∞ n=u an from n = u to n = v, we introduce a change of variable m = n+(v−u) and perform similar manipulations as shown above. In the example above, u = 1 and v = 3. Let r be a non-zero constant. You are asked to reindex the series P∞ n=1 1 rn−1 from n = 1 to n = 0. What is the result of reindexing? Select one or more: a. P∞ n=0 1 rn+1 b. P∞ n=0 1 rn−1 c. P∞ n=0 1 rn
Solution
3 0

Find the values of x for which the given geometric series converges. Also, find the sum of the series (as a function of x) for those values of x. ∑n = 0∞(−12)n(x−3)n Find the values of x for which the given geometric series converges. (Type an inequality or a compound inequality. Use integers or fractions for any numbers in the inequality. ) Find the sum of the series. ∑n = 0∞(−12)n(x−3)n =

Solution
0 0

Assuming Vtn = 1 V and Vx and Vy at time 0 = 0 V . ignoring the body effect find Vout at time t = infinity.
Solution
1 0