Question

Determining whether an infinite series converges or diverges

Original question: 2. If βˆ‘\sum and βˆ‘\sum

  1. Find the sum of βˆ‘n=0∞5n5n+5βˆ’n\sum_{n=0}^{\infty} \frac{5^n}{5^n + 5^{-n}} βˆ‘n=0∞5n5n+5βˆ’n\sum_{n=0}^{\infty} \frac{5^n}{5^n + 5^{-n}}

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Expert Verified Solution

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Expert intro: This problem asks for the behavior of an infinite series whose terms do not shrink to zero. The first step is to simplify the general term and test convergence before attempting any summation.

Detailed walkthrough

Rewrite the general term

The series shown is

βˆ‘n=0∞5n5n+5βˆ’n.\sum_{n=0}^{\infty} \frac{5^n}{5^n+5^{-n}}.

A useful algebraic simplification is to multiply the numerator and denominator by 5n5^n:

5n5n+5βˆ’n=52n52n+1.\frac{5^n}{5^n+5^{-n}}=\frac{5^{2n}}{5^{2n}+1}.

So the series becomes

βˆ‘n=0∞52n52n+1.\sum_{n=0}^{\infty} \frac{5^{2n}}{5^{2n}+1}.

Test the term behavior

For a series to converge, its terms must approach 00 as nβ†’βˆžn\to\infty. Here,

lim⁑nβ†’βˆž52n52n+1=1.\lim_{n\to\infty} \frac{5^{2n}}{5^{2n}+1}=1.

That limit is not zero. Therefore the necessary condition for convergence fails.

Conclude about the series

Because the terms do not approach zero, the infinite series

βˆ‘n=0∞5n5n+5βˆ’n\sum_{n=0}^{\infty} \frac{5^n}{5^n+5^{-n}}

does not converge. In fact, since each term is positive and approaches 1, the partial sums increase without bound, so the series diverges to infinity.

Why no summation formula applies

This is not a geometric series, p-series, or telescoping series. Even though the expression contains exponential terms, there is no constant ratio between consecutive terms of the series itself. The correct strategy is not to search for a closed-form sum, but to apply the divergence test first.

Key lesson

When you see an infinite series, always check the term limit before trying anything else. If the terms do not go to zero, the series cannot have a finite sum. Here the terms approach 1, which makes divergence immediate.

Final answer

TheΒ seriesΒ diverges.\boxed{\text{The series diverges.}}

πŸ’‘ Pitfall guide

A common mistake is to simplify the term correctly and then stop, assuming the presence of powers of 5 means the series must be geometric. It is not geometric because the denominator changes with nn in a non-constant way. Another mistake is to say the series converges because each term is less than 1. That is not enough: infinitely many positive terms that stay near 1 still produce unbounded growth. The zero-term test is the fastest and safest check here.

πŸ”„ Real-world variant

If the series were βˆ‘n=0∞5βˆ’n5n+5βˆ’n,\sum_{n=0}^{\infty} \frac{5^{-n}}{5^n+5^{-n}}, then multiplying top and bottom by 5n5^n would give 152n+1,\frac{1}{5^{2n}+1}, which does go to 0, so convergence would become possible and would need a different test. If the question instead asked for a finite sum such as βˆ‘n=0105n5n+5βˆ’n,\sum_{n=0}^{10} \frac{5^n}{5^n+5^{-n}}, then there is no convergence issue because only finitely many terms are involved. The distinction between finite and infinite sums is essential.

πŸ” Related terms

divergence test, term limit, infinite series

FAQ

How do you use the divergence test on an infinite series?

Compute the limit of the general term as n approaches infinity. If the terms do not approach zero, the series cannot converge.

Why does a term limit of one imply the series diverges?

An infinite series can only converge if its terms tend to zero. If the terms approach one, the partial sums keep growing, so the series diverges.

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