Solve/simplify This Series To A Normal Function/equation Of X

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Introduction

In mathematics, a series is a sum of terms that are defined by a formula. The given series is a classic example of an infinite series, where each term is defined by a specific formula. In this article, we will simplify the given series to a normal function or equation of xx. This will involve using various mathematical techniques, including algebraic manipulation and calculus.

Understanding the Series

The given series is:

∑n=2∞nnx\sum_{n=2}^{\infty} \frac{n}{n^x}

This series represents the sum of terms, where each term is defined by the formula nnx\frac{n}{n^x}. The series starts from n=2n=2 and goes to infinity.

Breaking Down the Series

To simplify the series, we need to break it down into smaller parts. We can start by rewriting the series as:

∑n=2∞nnx=∑n=2∞1nx−1\sum_{n=2}^{\infty} \frac{n}{n^x} = \sum_{n=2}^{\infty} \frac{1}{n^{x-1}}

This simplification involves using the property of exponents, where 1nx=1nx−1\frac{1}{n^x} = \frac{1}{n^{x-1}}.

Using the Formula for the Sum of a Geometric Series

The series we have now is a geometric series, where each term is defined by the formula 1nx−1\frac{1}{n^{x-1}}. The sum of a geometric series can be calculated using the formula:

∑n=0∞arn=a1−r\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}

where aa is the first term and rr is the common ratio.

Applying the Formula

In our case, the first term is 12x−1\frac{1}{2^{x-1}} and the common ratio is 12x−1\frac{1}{2^{x-1}}. Applying the formula, we get:

∑n=2∞1nx−1=12x−11−12x−1\sum_{n=2}^{\infty} \frac{1}{n^{x-1}} = \frac{\frac{1}{2^{x-1}}}{1-\frac{1}{2^{x-1}}}

Simplifying the Expression

We can simplify the expression further by using algebraic manipulation. We can rewrite the expression as:

12x−11−12x−1=12x−1−1\frac{\frac{1}{2^{x-1}}}{1-\frac{1}{2^{x-1}}} = \frac{1}{2^{x-1}-1}

Conclusion

In this article, we simplified the given series to a normal function or equation of xx. We used various mathematical techniques, including algebraic manipulation and calculus, to break down the series into smaller parts. The final expression is:

12x−1−1\frac{1}{2^{x-1}-1}

This expression represents the simplified series, where each term is defined by the formula 1nx−1\frac{1}{n^{x-1}}. The series starts from n=2n=2 and goes to infinity.

Final Answer

The final answer is 12x−1−1\boxed{\frac{1}{2^{x-1}-1}}.

Additional Information

For more information on sequences and series, please visit the following resources:

References

Related Topics

Further Reading

For more information on mathematical techniques, please visit the following resources:

Introduction

In our previous article, we simplified the given series to a normal function or equation of xx. In this article, we will answer some frequently asked questions related to the series and its simplification.

Q: What is the given series?

A: The given series is:

∑n=2∞nnx\sum_{n=2}^{\infty} \frac{n}{n^x}

This series represents the sum of terms, where each term is defined by the formula nnx\frac{n}{n^x}. The series starts from n=2n=2 and goes to infinity.

Q: How did you simplify the series?

A: We simplified the series by breaking it down into smaller parts. We started by rewriting the series as:

∑n=2∞nnx=∑n=2∞1nx−1\sum_{n=2}^{\infty} \frac{n}{n^x} = \sum_{n=2}^{\infty} \frac{1}{n^{x-1}}

This simplification involved using the property of exponents, where 1nx=1nx−1\frac{1}{n^x} = \frac{1}{n^{x-1}}.

Q: What is the formula for the sum of a geometric series?

A: The formula for the sum of a geometric series is:

∑n=0∞arn=a1−r\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}

where aa is the first term and rr is the common ratio.

Q: How did you apply the formula to the series?

A: We applied the formula to the series by identifying the first term and the common ratio. In our case, the first term is 12x−1\frac{1}{2^{x-1}} and the common ratio is 12x−1\frac{1}{2^{x-1}}. Applying the formula, we get:

∑n=2∞1nx−1=12x−11−12x−1\sum_{n=2}^{\infty} \frac{1}{n^{x-1}} = \frac{\frac{1}{2^{x-1}}}{1-\frac{1}{2^{x-1}}}

Q: How did you simplify the expression further?

A: We simplified the expression further by using algebraic manipulation. We can rewrite the expression as:

12x−11−12x−1=12x−1−1\frac{\frac{1}{2^{x-1}}}{1-\frac{1}{2^{x-1}}} = \frac{1}{2^{x-1}-1}

Q: What is the final expression for the series?

A: The final expression for the series is:

12x−1−1\frac{1}{2^{x-1}-1}

This expression represents the simplified series, where each term is defined by the formula 1nx−1\frac{1}{n^{x-1}}. The series starts from n=2n=2 and goes to infinity.

Q: What are some related topics to this article?

A: Some related topics to this article include:

Q: Where can I find more information on sequences and series?

A: For more information on sequences and series, please visit the following resources:

Conclusion

In this article, we answered some frequently asked questions related to the series and its simplification. We hope this article has provided you with a better understanding of the series and its simplification.

Final Answer

The final answer is 12x−1−1\boxed{\frac{1}{2^{x-1}-1}}.

Additional Information

For more information on sequences and series, please visit the following resources:

References

Related Topics

Further Reading

For more information on mathematical techniques, please visit the following resources: